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SD 9-12 Algebra03/20/2009

Student Name: ___________________________

Class: ___________________________

Date: ___________________________

Instructions: Read each question carefully and select the correct answer.

1. The number of mosquitos in Kylie's backyard varies directly as the amount of water in a puddle. When there are 4 gallons of water in the puddle, there are 120 mosquitos in the yard. How many mosquitos would be in the yard if there was 1 gallon of water in the puddle?

A. 116 mosquitosB. 480 mosquitosC. 30 mosquitosD. 481 mosquitos

2. Calculate the slope of the line between the points (- 4, 10) and (3, 8).

A. 2

B.

C.

D.

SD Algbra-Questions, Answer Key, Study Guide of 104

3. Factor completely.

4x2 - 36

A. (2x + 6)(2x - 6)B. 2(x + 9)(x - 16)C. (4x + 18)(x - 2)D. 4(x + 3)(x - 3)

4. Simplify.

A.

B.

C.

D.

5. A riding stable charges $150 per month to board a horse plus $6.75 per bale of hay that the horse eats. What is the maximum number of hay bales that Shiloh eats in a month if his monthly boarding cost is at least $225.00 and at most $255.00? Round your answer to the nearest whole bale that satisfies the inequality.

A. 12 balesB. 11 balesC. 16 balesD. 15 bales

6. The electrical resistance (R) of a wire varies directly with its length (L) in meters, and indirectly with the square of its diameter (d) in millimeters. The equation that represents this relationship is given by

where k is the constant for wire made out of a specific metal. If the resistance of a 50 meter long wire is 8.64 ohms when its diameter is 5 millimeters, what is the resistance of a 200 meter long wire when its diameter is 8 millimeters?

A. 13.5 ohmsB. 14.3 ohmsC. 15.4 ohmsD. 17.9 ohms

7. Find the range of the following relation r.

A.

B.

C.

D.


8. Which function rule relates the values of the input variable, x, to the values of the output variable, y, in the table below?

A. y = x - 4B. y = x + 5C. y = x + 7D. y = x - 1

9.

A. QB. RC. SD. T

10. Simplify.

A. 3B. 9C. 324

D.

11. Subtract the following.

A.

B.

C.

D.

12. Evaluate the expression with y = 18.

3 - 2(4y - 5)

A. -131B. -151C. 137D. 157

13. Two sides of a rectangular picture frame are each 7 inches shorter than the other two sides. The perimeter of the picture frame measures 88 inches. What is the length of the shortest two sides of the frame?

A. 25.5 in.B. 30 in.C. 18.5 in.D. 37 in.


14. Salim had a square picture that he enlarged. He increased the length by 3 inches and the width by 1 inch so that the area of the picture is 35 in2 . What was the length of the picture's sides before he enlarged it?

A. 31 inB. 5 inC. 4 in.D. 7 in.

15. Which of the following points, if removed from the set, would make the set a function?

A. (- 9, 9)B. (5, - 9)C. (- 5, 5)D. (9, - 5)

16. Solve for k.kl - 4 = 2k

A.

B.

C.

D.

17. Solve for h.

A.

B.

C.

D.

18. Find the missing number.

9 + (4 + 7) = 4 + (9 + ?)

A. 9B. 4C. 7D. 20

19. Solve for the value of x.

A. x = 36B. 6 < x < 7C. x = 6D. 5 < x < 6

20. To get a certain shade of orange, Raul needs to mix red paint with yellow paint in the ratio of 5:2.

Which of the following equations shows how many quarts of red paint Raul needs to mix with 4 quarts yellow paint?

A. 5/2 = x/4B. 5/2 = 4/xC. 5/8 = xD. 8/5 = x


21. Complete the following linear function table.

A. 3B. 4C. 1.5D. 4.5

22. What is the value of x in the given statement?

A.

B.

C.

D.

23. If f(x) = 0.5x + 3.1 and g(x) = 0.6x - 1.86, find g(f(x)).

A. 0.3x2 + 0.93x - 5.766B. 1.1x + 1.24C. 0.3x + 2.17D. 0.3x

24. Write the absolute value function, f(x) = |-3x - 4|, as a compound function.

A.

B.

C.

D.

25. Find the area of an ellipse with a major axis of 8 in. and a minor axis of 2 in. Express your answer in terms of .

A. 8 in.2

B. 64 in.2

C. 4 in.2

D. 16 in.2

26. Solve by completing the square.

x2 - 3x - 21 = 0

A. x = - 7 or x = 3

B.

C.

D. No Real Roots


27. Solve for x.

A. 1B. -128C. -2D. -8

28. What is the value of y?

y + 8 - 11 = 17

A. -2B. 20C. 46D. -2

29. What is the value of x?

x0.2 = -- 15

A. 7.5B. 75C. 3D. 3.3

30. Solve the inequality.

n + 8 < -7

A. n > -1B. n < -15C. n > -15D. n < -1

31. What is the value of n?

5n + 15 = 100

A. n = 575B. n = 425C. n = 23D. n = 17

32. What is the value of y?

x = 9yz = 10y

A. y = 90B. y = 1 1/9C. y = 1D. Problem cannot be solved with the given information.

33. Which of the following is true?

A.

B.

C.

D.

34. Find the missing symbol.

(8 x 8) ? (7 x 7) = 15

A. +B. -C. xD. ÷


35. Round to the nearest cent when necessary.

Which of the following is the best price?

A. 8 for $3.99B. 7 for $3.55C. 6 for $3.49D. 9 for $4.20

36. Which of the following best completes the number sentence?

2 = (18 x 2) + (14 x 2)

A. (18 x 14) +B. (18 x 14) xC. (18 + 14) xD. (18 + 14) +

37. Simplify

A. AB. BC. CD. D

38. Which relation is written as a function?

A. x equals negative y squaredB. y equals negative x to the fourth powerC. x equals negative y to the second powerD. x equals 3y squared

39. Four pancakes and three eggs at Candy's Cafe cost $7.95. Two pancakes and three eggs at Burger Palace cost $5.95. Which option shows the best method for calculating the amount that each restaurant is charging for each pancake and each egg?

A. -2p = $13.90B. 6p = $13.90C. 4p + 3e = $7.95 and 2p + 3e = $5.95D. 4 + 3(p + e) = $7.95 and 2 + 3(p + e) = $5.95

40. Given the points O(-3, 4), Q(-2, 6), write an equation for the line OQ and find the y-coordinate for point P (3, y) on the line OQ.

A. P(3, 2)B. P(3, 4)C. P(3, 44/5)D. P(3, 16)

41. The length of a rectangle is four more than three times the width. The perimeter is sixty-four meters. Find the length and the width.

A. l = 21.5; w = 10.5B. l = 25; w = 7C. l = 26.5; w = 5.5D. l = 28; w = 4

42. Factor completely.

6x2 - 3x - 30

A. 3(2x - 5)(x + 2)B. 3(2x - 2)(x + 5)C. 3(2x + 5)(x - 2)D. 3(2x + 2)(x - 5)


43. Solve the following using the quadratic formula.

- 2x2 + 3x + 4 = 0

A.

B.

C.

D.

44.

A.

B.

C.

D.

45. Simplify.

A. AB. BC. CD. D

46. Simplify.

(a + b)(3a + 2b)

A.

B.

C.

D.

47. Simplify.

A. AB. BC. CD. D

48. Find the solution set and choose the corresponding description: 12 - (10 - 8 - 3x) < 8 - 10 + 2x

A. all points to the left of -2B. all points to the left of -12, including -12C. all points to the left of -12D. all points to the left of -2, including -2

49. Solve for x.

|8x + 12| - 13 = 23

A. x = 6 or x = - 3B. x = 6C. x = 3D. x = 3 or x = - 6


SD 9-12 AlgebraAnswer Key03/20/2009

1. C Direct Variation 2. D Slope 3. D Factoring: Difference of Two Squares 4. C Radicals: Addition/Subtraction - A 5. D Inequalities - C 6. A Joint and Combined Variation 7. A Domain/Range 8. C Function Rules 9. A Exponential Functions 10. A Exponential Notation - F 11. A Polynomials: Subtraction 12. A Equations: Order of Operations 13. C Functions: Linear 14. C Quadratic Equations: Real World Problems 15. A Functions/Relations - A 16. A Equations With Two Variables 17. D Literal Equations 18. C Missing Elements - D 19. C Radicals and Roots 20. A Ratio/Proportion - C 21. B Function/Pattern - C

22. D Inequalities - A 23. D Functions: Quadratic 24. B Functions: Absolute Value 25. C Irrational Numbers: Pi 26. B Non-Linear Equations 27. B Rational Numbers: Equations 28. B Equations: Addition/Subtraction 29. C Equations: Multiplication/Division 30. B Inequalities - B 31. D Equations: Two-Step 32. D Solving Equations: Substitution 33. C Exponential Notation - D 34. B Missing Elements - E 35. D Rates 36. C Properties - E 37. A Radicals: Simplifying 38. B Functions/Relations - B 39. C Equations: Systems 40. D Equations of a Line 41. B Number Relation Problems 42. A Factoring 43. B Quadratic Formula 44. D Polynomials: Addition 45. C Exponential Notation - E 46. C Polynomials: Multiplication 47. B Polynomials: Division 48. C Sets/Subsets/Solution Sets 49. D Absolute Value: Solve


Study Guide

SD 9-12 Algebra03/20/2009

Direct VariationVariation equations are formulas that show how one quantity changes in relation to one or more other quantities. There are four types of variation: direct, indirect (or inverse), joint, and combined.

Direct variation equations show a relationship between two quantities such that when one quantity increases, the other also increases, and when one quantity decreases, the other also decreases. We can say that y varies directly as x, or y is proportional to x. Direct variation formulas are of the form y = kx, where the number represented by k does not change and is called a constant of variation.

Indirect variation equations are of the form y = k/x and show a relationship between two quantities such that when one quantity increases, the other decreases, and vice versa.

This skill focuses on direct variation. The following is an example of a direct variation problem.

The amount of money in a paycheck, P, varies directly as the number of hours, h, that are worked. In this case, the constant k is the hourly wage, and the formula is written P = kh. If the equation is solved for k, the resulting equation shows that P and h are proportional to each other.

Therefore, when two variables show a direct variation relationship, they are proportional to each other. Direct variation problems can be solved by setting up a proportion in the form below.

Example 1: The amount of fuel needed to run a textile machine varies directly as the number of hours the machine is running. If the machine required 8 gallons of fuel to run for 24 hours, how many gallons of fuel were needed to run the machine for 72 hours? Round your answer to the nearest tenth of a gallon, if necessary.


Step 1: Set up the proportion. Since the machine used 8 gallons of fuel in 24 hours, the left side of the proportion should be 8 gallons over 24 hours. The number of gallons that the machine used in 72 hours needs to be found, so the right side of the proportion should be g gallons over 72 hours.Step 2: Cross-multiply across the equal sign.Step 3: Set up the cross-multiplication equation.Step 4: Divide both sides of the equation by 24 hours to isolate g gallons.Step 5: Reduce the fractions on both sides of the equal sign.Step 6: Simplify by multiplying the numbers remaining on the right side of the equal sign (8 gallons 3).

Answer: 24 gallons

Example 2: The price of jellybeans, j, varies directly as the number of pounds, p, that are purchased. Find the equation that relates the two variables if jellybeans are $1.95 per pound.

(1) y = kx, j = kp(2) j = 1.95p

Step 1: Remember that the formula for direct variation is: y = kx and substitute the variables from the question into the appropriate places.Step 2: Since the jellybeans are always $1.95 per pound, the constant, k, equals 1.95. Substitute 1.95 into the equation for k.

Answer: j = 1.95p

Activities that can help reinforce the concept of direct variation are as follows.1. Have students solve the equation y = kx for k, and then substitute two sets of (x, y) values into

the equation and compare the values for k. If they are the same, then x and y have a direct variation relationship.

2. Have the student think of scenarios that show a direct variation relationship. Then, make up numbers to go with the relationships and have the students practice solving them.

SlopeSlope is the ratio of the vertical difference of two points on a line and the horizontal difference between the same two points. Slope is also defined as "rise over run" and is found by calculating the difference in the y-


coordinates (rise) divided by the difference in the x-coordinates (run). Slope can be found either graphically or algebraically.

Finding Slope Graphically:

Two points are identified on line z: A (- 3, 8) and B (3, 1).To find the slope of line z, which passes through points A and B, follow these steps.

Step 1: Start at the left-most point (it is possible to start at either point, but it is important to be consistent), which is point A (- 3, 8).Step 2: Count up (positive) or down (negative) until level with the right-most point, point B (3, 1). The count will be 7 in the downward direction, or - 7. The rise is - 7.Step 3: Count right (positive) or left (negative) until point B is reached. The count will be 6 to the right, or +6. The run is 6.Step 4: Using the "rise over run" definition of slope, place - 7 on top of the fraction and 6 on the bottom of the fraction.

Finding Slope Algebraically:To find the slope of line z algebraically, follow these steps.

Step 1: Use the formula for the slope of a line, where m is the variable that represents slope.

Step 2: Step 3: Substitute the given coordinate points into the slope formula and simplify the fraction.

It does not matter which method (graphically or algebraically) is used for determining slope. Regardless, the slope of a line always remains the same.

Example 1: Find the slope of the line between Point R (2, 4) and Point S (1, 3).


Graphically:Step 1: Start at the left-most point, which is point S (1, 3).Step 2: Determine the rise (1).Step 3: Determine the run (1).Step 4: Using the "rise over run" definition of slope, place 1 on top of the fraction and 1 on the bottom of the fraction.

Answer: m = 1

Algebraically:

Step 1: Write the formula. Step 2: Step 3: Simplify the fraction.

Answer: m = 1

An activity that can reinforce the concept of slope is to have students randomly plot two points on a coordinate system and then find the slope graphically. They can check their answers by substituting the two points into the slope formula.

Factoring: Difference of Two SquaresConsider the following equation:

3 4 = 12The numbers 3 and 4 are said to be factors of the number 12. Factors are terms that can be multiplied together to create another term. This concept of factoring is not reserved for numbers, but may be extended to polynomials as well.

Definitions:• Factoring is rewriting a mathematical expression as a product of its component factors. One way to think of factoring is as the opposite or inverse of multiplying.• Sum means the addition of two terms.• Difference means the subtraction of two terms.• Variables are letters that represent unknown numbers.• Squares are terms whose square roots are whole numbers, zero, or variables whose powers are


divisible by two. For example, 81 is a square because the square root of 81 is 9, and x2 is a square because the exponent (2) is divisible by two.• A polynomial is a term or sum/difference of terms. Each term is either a number, a variable, or a product of a number and one or more variables. Examples: 3x2 + 2x +1; 5x3 ; 4x + 1• A monomial is a polynomial with one term. Example: 6x2 • A binomial is a polynomial with two terms. Example: 8y4 + 7y • A trinomial is a polynomial with three terms. Example: 3x2 + 2x +1

The difference of two squares is easily factored using the rule below.a2 - b2 = (a - b) (a + b)

Example 1: Factor completely.x2 - 169

Step 1: This problem is a difference of two squares since x2 and 169 are perfect squares. Set up two sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of the other set.Step 2: Take the square root of the first term (x2 ) and write x in the first space in each set of parentheses.Step 3: Take the square root of the second term (169) and write 13 in the second space in each set of parentheses.

Answer: (x - 13)(x + 13)

NOTE: This format will always work because if (x - 13) and (x + 13) are multiplied together using the FOIL method, (multiply the First terms, then the Outer terms, then the Inner terms, then the Last terms), the result is: x2 + 13x - 13x - 169. Combine like terms, or terms that have the same variable(s) with the same exponent(s), the +13x and -13x terms add to zero. The result is x2 - 169, which is the original problem.

NOTE: Because of the commutative property of multiplication, the order of the factors can be reversed and an equivalent expression is made: (x - 13)(x + 13) = (x + 13)(x - 13).

Example 2: Factor completely.169y2 - 36

Step 1: This problem is a difference of two squares since 169y2 and 36 are perfect squares. Set up two sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of the other set.Step 2: Take the square root of the first term (169y2 ) and write 13y in the first space in each set of parentheses.


Step 3: Take the square root of the second term (36) and write 6 in the second space in each set of parentheses.

Example 3: Factor completely.2x2 - 50

Step 1: Factor out any common factors. Since 2x2 and 50 can both be evenly divided by 2, perform the division. NOTE: Don't forget to carry the 2 that was factored out through the entire problem.Step 2: The remaining problem is a difference of two squares since x2 and 25 are perfect squares. Set up two sets of parentheses, and then write a minus sign in the center of one set, and a plus sign in the center of the other set.Step 3: Take the square root of the first term (x2 ) and write x in the first space in each set of parentheses.Step 4; Take the square root of the second term (25) and write 5 in the second space in each set of parentheses.

Answer: 2(x - 5)(x + 5)

To help the student further understand this concept, write square numbers on one set of index cards. Write squared variable terms and/or the product of numbers and variables on another set of index cards. Have the student draw one card from each set of index cards. Ask the student to factor the differences between each pair of cards. He or she will have to find the square roots of the terms on the cards to correctly complete the task.

Radicals: Addition/Subtraction - AA radical sign looks like a check mark with a line across the top. The radical sign is used to communicate square roots. The square root of a number is the number that when squared, or multiplied by itself, is equal to the radicand. The radicand is the number under the radical sign. For example, the square root of 64 is equal to 8, because 8 squared (8 8) is 64, and 64 is the radicand.

Students must be able to combine (add and subtract) radicals. In order for radicals to be combined, the radicands must be identical. The square root of 3 and the square root of 7 cannot be added together in their present form because their radicands (in this case, 3 and 7) are not identical. When combining radicals, the radicands themselves do not get added or subtracted.


Example 1: Add and simplify.

Step 1: Check to see if the radicands are identical. In this case, both radicands are 5, so the radicals can be added.Step 2: Add the coefficients (numbers in front of the radicals). The answer is the sum of the coefficients times the radical.

Example 2: Subtract and simplify.

Step 1: Check to see if the radicands are identical. In this case, both radicands are 3, so the radicals can be subtracted.Step 2: Subtract the coefficients and the answer is the difference of the coefficients times the radical. There is a silent 1 in front of the radical in the final answer.

Example 3: Simplify.

Step 1: Check to see if the radicands are identical. In this case, two of the radicands are 7, so those two radicals can be subtracted.Step 2: Subtract the coefficients and the answer is the difference of the coefficients times the radical, plus the remaining radical.


Step 1: Check to see if the radicands are identical. In this case, two of the radicands are 3, so these two SD Algbra-Questions, Answer Key, Study Guide of 104

radicals can be added. Add the coefficients and then multiply the sum by the radical.Step 2: Rewrite radical 16 in simplest form. Since 16 = 4 4, radical 16 can be simplified to 4.Step 3: The final answer is the answer to step 1 minus the answer to step 2.

An activity that can help reinforce the concept of adding and subtracting radicals is to name radicals with the names of animals. For example, name the square root of 6 "elephant," and the square root of 5 "lion." Then say, "If you have 5 elephants and you add 10 more elephants and 2 lions, how many elephants and lions do you have altogether?" Answer: 15 elephants and 2 lions, or, 15 times the square root of 6 plus 2 times the square root of 5. This activity keeps students from trying to combine radicands in the answer.

Inequalities - CAn inequality is a number sentence that uses "is greater than," "is greater than or equal to," "is less than," "is less than or equal to," or "is not equal to" symbols. For example, 6n > 4 is a number sentence with an inequality symbol, and is read "six times n is greater than four." Inequalities can be identified in real world situations by expressions such as "is less than," "is more than," "at least," and "at most." Sentences can be translated to number sentences using the following symbols.

An inequality can be solved for a variable (a letter that represents a number) in the same way that an equation is solved. An example of solving an inequality follows.

Example 1: Solve the inequality for x.2x + 7 > 11

Step 1: Subtract 7 from each side of the inequality symbol to isolate the variable term, 2x, and simplify.Step 2: Divide both sides of the inequality by 2 to completely isolate x.

Answer: x > 2

A compound inequality has more than one condition and can be identified in a number sentence by two inequality symbols, or in a real world application by the words "and" or "or." For example, 3 < x < 9 is a compound inequality. To read a compound inequality, start in the center and read left: x > 3, then go back to the center and read to the right: x < 9. The compound inequality is a combination of x > 3 AND x < 9. Solving a compound inequality is very similar to solving a single inequality.Example 2: Solve the compound inequality for x.

4 < 3x + 1 < 7SD Algbra-Questions, Answer Key, Study Guide of 104

Step 1: Subtract 1 from the left, center, and right side of the number sentence to isolate the variable term, 3x.Step 2: Divide the left, center, and right side of the number sentence by 3 to isolate x.

Answer: 1 < x < 2. This answer can be interpreted to mean that x is greater than 1 and x is less than 2.

Although the above examples do not include negative numbers, students must remember to switch the inequality sign when multiplying or dividing by a negative number. Once the student is familiar with solving compound inequalities, he or she should be ready to solve them in real world situations.

Example 3: The telephone company charges $21.95 per month for basic service plus $0.17 per local call. What is the maximum number of calls that can be made in a month if a family can spend at least $27.50 and at most $32.75 per month for telephone service? Round your answer to the nearest whole number of calls that satisfies the inequality.

Step 1: Translate the information in the problem to a compound inequality. Let c represent the number of calls. The family gets charged $21.95 + $0.17c each month (the service charge plus local calls). Since the charges are at least $27.50 and at most $32.75, place the minimum amount of $27.50 to the left of the expression that represents the charges and the maximum amount of $32.75 to the right of the expression. Refer to the chart above to see which inequality symbol to use. Since the question uses the words at least and at most, the needs to be used. Place the on either side of the charges expression.Step 2: Isolate the variable term $0.17c by subtracting $21.95 from the left, center, and right of the compound inequality. Simplify.Step 3: Divide each term of the new compound inequality by $0.17 to isolate c.Step 4: Simplify the inequality.Step 5: Since the maximum number of calls must be less than or equal to 63.53, the whole number that will satisfy the inequality is 63. Rounding up to 64 would no longer satisfy the inequality because 64 is greater than 63.53.

Answer: 63 calls

Note: If the question had asked what the minimum number of calls would be, the answer would be 33,


since the minimum number of calls has to be greater than or equal to 32.65.

An activity that can help reinforce the concept of inequalities is to ask the student to make a budget using the amount of money he or she earns each month (approximate or make up an amount if necessary). Then, make up scenarios using the budget. For example:It costs $120 per month to insure a car, plus $2.05 per gallon of gasoline used. If you have budgeted at least $150.00 and at most $200.00 for car expenses this month, what is the maximum number of gallons of gas you can buy?

Joint and Combined VariationVariation equations are formulas that show how one quantity changes in relation to one or more quantities. There are four types of variation: direct, indirect (or inverse), joint, and combined. This skill focuses on joint and combined variation.

Direct variation equations show a relationship between two quantities such that when one quantity increases, the other also increases, and vise versa. We can say that y varies directly as x. Direct variation formulas are of the form y = kx, where the number represented by k does not change and is called a constant of variation. For example, the amount of money in a paycheck (P) varies directly as the number of hours (h) worked. In this case, the constant k is the hourly wage, and the formula is written P = kh.

Indirect variation formulas show that when one quantity increases, the other quantity decreases, and vise versa. For example, when the price of an item increases, the demand decreases. Indirect variation formulas are of the form y = k/x.

Joint variation formulas show the relationship between a quantity and the product of two other quantities. They are of the form y = kxz. For example, the equation for the area of a triangle states that the area is equal to one half times base times height, or A = ? bh. The number ? is the constant of variation, and is always the same for the area of a triangle.

Combined variation formulas show how a quantity varies directly with one or more quantities and indirectly with one or more different quantities. They are of the form y = (kx)/z. For example, the sales (S) of a company vary directly as the amount spent on advertising (a), and indirectly as the price (p) of the item sold. The formula is written as S = (ka)/p. When the amount spent on advertising increases, sales increase because sales and advertising vary directly. When the price increases, sales decrease, because price and sales vary indirectly.

To solve variation problems, follow these steps.

Step 1: Substitute the quantities given for the letters they represent. Step 2: Solve for the constant k. Step 3: Set up the equation again. Step 4: Substitute given quantities and the value of k that was just determined. Step 5: Solve for the quantity needed.

Example 1: The surface area (S) of a cube varies jointly with the length (l) and width (w) of one of the faces. The equation that represents this relationship is given by


S = klw

where k is a constant that needs to be determined. If the surface area of a cube is 150 square inches when the length and width of each face are 5 inches, what are the length and width of each face when the surface area is 216 square inches?

Step 1: Write the values for the variables that are known (do not use the values for a surface area of 216 because the constant k is not known).Step 2: Substitute the values into the formula and solve for k. Multiply the terms on the right side of the equal sign to get 25k. Isolate the k by dividing each side of the equation by 25. k = 6.Step 3: Now that the constant is known, it is possible to determine the length and width of a cube that has a surface area of 216 in.2 . Write the values for the variables that are known for a surface area of 216.Step 4: Substitute the values into the variation equation and solve for l and w. Multiply the right side of the equation to get 6l2 . Since length and width are the same for a cube, (l)(w) can be written l2 or w2 . Divide each side of the equation by 6 to isolate the l2 . To remove the square from l2 , take the square root of each side of the equation. l = 6 and w = 6 (because l and w are the same).Answer: length = 6 inches and width = 6 inches

Example 2: The amount of monthly sales (S) for a company that makes a certain type of chair varies directly with the amount spent on advertising each month (a) and indirectly with the price (p) of the chair. The equation that represents this relationship is given by

where k is a constant to be determined. If the sales for July were $40,000.00 when advertising costs were $100.00 and the price of the chair was $50.00, what were the sales for August if advertising costs were increased to $135.00 and the price of the chair was decreased to $45.00?


Step 1: Write the values for the variables that are known (do not use the values for advertising cost of $135 and price of $45 because the constant k is not known).Step 2: Substitute the values into the formula and solve for k. Multiply the terms on the top of the fraction to get 100k. Divide 100k by 50 to further simplify the right side of the equation and get 2k. Isolate the k by dividing both sides of the equation by 2, yielding k = 20,000.Step 3: Now that the constant is known, it is possible to determine the sales for advertising costs of $135 and price of $45. Write the values for the variables that are known for these constraints.Step 4: Substitute the values into the variation equation and solve for S. Multiply the numbers on the top of the fraction to get 2,7000,000. Divide 2,700,000 by 45 to continue to simplify the equation and determine that the sales were $60,000.00Answer: $60,000.00

An activity that can help reinforce the concept of variation is to describe situations involving joint and combined variation, and then have the student write the equations relating the quantities. For example, "The volume of a cylinder varies jointly as the square of the radius of the cylinder and its height." The student would write: V = kr2 h.

Domain/RangeA relation is a collection of (x, y) coordinate points. For example:

The domain of the relation is the set of x-values in the collection of coordinate points that make up the relation, and the range is the set of y-values. The domain and range of a relation are written as sets of numbers in ascending order within brackets and each number is only written once. For example:

This skill focuses on giving students practice determining the domain and range of relations.

Example 1: Find the domain of the following relation r.


Solution: The domain is the set of all x-values of the coordinate points that make up the relation.

Answer: Example 2: Find the range of the following relation r.

Solution: The range is the set of all y-values of the coordinate points that make up the relation. NOTE: Although the - 4 is used as a y-value twice, it only needs to be written once in the range.

Answer: An activity that can help reinforce the concept of domain and range is to show students a collection of coordinate points and ask them to name the member of the domain of each coordinate point (x-value) and the member of the range of each coordinate point (y-value).

Function Rules

A function rule is an equation relating two variables. Variables are letters that represent unknown numbers, such as x and y.

In a function, the value of the output variable, y, depends on the value of the input variable, x. The function rule takes a given value of x, and performs an operation (addition, subtraction, multiplication, or division) on it to make it equal to y.

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A table of related values for input-output variables that can be generated from a function rule is called a function table. It is important for students to understand what a function table represents. To create the function table below, x-values were chosen and placed in the table. Next, the x-values were substituted into the function rule and a y-value was generated for each x-value. The final function table appears on the right without the function rule.


It is possible to determine the function rule from a table of values for related input-output variables. The first two examples deal with simple function rules that use addition and subtraction only.

Example 1: What function rule relates the values of the input variable, x, to the values of the output variable, y, in the table below?

Step 1: Determine what needs to be added to or subtracted from - 3 to make it equal to 3. The solution is, "add 6," because - 3 + 6 = 3.Step 2: Add 6 to each of the remaining x-values, and verify that they match the values listed for y.Step 3: Write the function rule: y = x + 6.


Step 1: Determine what needs to be added to or subtracted from 1 to make it equal to - 4. The solution is, "subtract 5," because 1 - 5 = - 4.Step 2: Subtract 5 from each of the remaining x-values, and verify that they match the values listed for y.Step 3: Write the function rule: y = x - 5.

Function rules can also be of the form y = ax + b, or y = ax. Rules of this form require the student to use multiplication, as well as addition and subtraction.



Step 1: First, the student should test the y values to see if they are multiples of the x values. If the function is of the form y = ax, then the student should be able to multiply all of the values in the x column by the same number, and arrive at the values in the y column. Multiplying - 4 by 3 yields - 12. Step 2: Multiply the remaining x values by 3, and verify that they match the values listed for y.Step 3: Write the function rule: y = 3x.


Step 1: First, the student should test the y values to see if they are multiples of the x values. If the function is of the form y = ax, then the student should be able to multiply all of the values in the x column by the same number, and arrive at the values in the y column. If the student is unable to determine a single factor that yields the output values, he or she should continue to step 2. Since there isn't a factor that yields - 19 when multiplied by - 4, the student should continue to step 2. Step 2: If there is an input value equal to 0, the student should always begin with this value. This line in the table will help the student to determine the value that is being added or subtracted since any factor multiplied by 0 equals zero. In this case, 3 is being subtracted from x because 0 - 3 = - 3. Therefore, the function must be of the form y = ax - 3.Step 3: The student must now determine what value of a satisfies the function. Determining this value is made easier by performing the inverse of the operation found in step 2 on a chosen y value. Adding 3 to -19 yields - 16. Since - 4, (the corresponding x value) must be multiply by 4 in order to yield -16, the factor 4 is the correct value for a. Step 4: Multiply each input value by 4 and subtract 3 in order to verify that they match the values listed for y.Step 5: Write the function rule: y = 4x - 3.


Exponential FunctionsAn exponential function is an equation that has a variable in the exponent. The functions below are all exponential functions, where f(x) represents the letter y and is dependent on the value of x.

When x-values are substituted into the equation and the right side of the equal sign is simplified, a list of f(x) (or y) values is obtained. This process gives you a set of (x, y) coordinate pairs that can be plotted (or graphed) on a coordinate plane.

Step 1: Make a table of values with the following column headings.

Step 2: Choose values to enter into the x column (remember to choose the smallest possible values to keep computations as simple as possible). It is generally necessary for the values to contain both negative and positive numbers and the number zero. As few as three values could be used, but this example will use these five values: -2, -1, 0, 1, 2. Enter the values into the table.Step 3: Substitute each x-value into the equation in the center column for x and simplify. This will determine the y-values for the coordinate points.


Step 4: Write the (x, y) pairs generated in Step 3 into the (x, y) column, using the values from the x column and the y-values from the center column.

Step 5: Plot the points on a coordinate plane to graph the exponential function.

Step 6: Compare this graph to the solution choices and choose the correct answer, which is choice D.

Answer: D.

Step 1: Make a table of values and choose x-values to substitute into the equation.

Step 2: Substitute the values in the x column into the equation in the center column to determine the corresponding y-values. Write the coordinate points


in the (x, y) column.

Step 3: Plot the points on a coordinate plane to graph the exponential function.

Step 4: Compare this graph to the curves in the question and choose the correct answer.

Answer: W

Comparing Graphs of Exponential Functions:The standard form for an exponential function is shown below.

For purposes of comparing graphs of exponential functions, two more variables need to be added, such that the standard form becomes the form shown below.

The value of c determines whether the graph shifts upward or downward and the value of d determines whether the graph shifts right or left. See the table below.

Example 3: Step 1: Step 2: Look at the table above to determine that the graph of the exponential function should move upward 3 units and to the right 4 units.Step 3: Compare the actual graphs below to see if these predictions were right.

An activity that will help reinforce the skill of graphing exponential functions is to explore exponential


functions with the student using a graphing calculator. Make up a list of exponential functions like the one below.

Follow the steps to graph each function using a graphing calculator. Notice the way changing a sign or adding a term alters the graph.

Step 1: Follow the instructions in your graphing calculator manual to set the viewing window to the following values: x min = -10, x max = 10, xscl = 1; y min = -10, y max = 10, y scl = 1. Exit the window.Step 2: Press the y = key to open the equation editor. Delete any equations that are listed. Enter the following keystrokes for the first exponential equation on the list: 5^x. Press Enter.Step 3: Press the Graph key to display the graph of the function.

**NOTE: This activity could be completed using graph paper and tables of values to graph the exponential functions.

Exponential Notation - FIn the expression 32 , the number 2 is called an exponent and the number 3 is called a base. The exponent determines the number of times the base is multiplied by itself. For example: 32 = (3)(3) = 9.

Rational exponents are exponents that are in fraction form. The following are examples of expressions with rational exponents.

Expressions with rational exponents can be written in radical form. The numerator is the number of times the base, which is placed under the radical sign, is multiplied by itself. The denominator is the root of the radical expression. A root is the inverse of an exponent. The examples below show how expressions with rational exponents are written in radical form.

A root of 2 is called a square root and is the inverse of the exponent 2. A root of 3 is called a cube root and is the inverse of the exponent 3. The following examples show how to simplify expressions with rational exponents in two ways.

Simplifying by converting to radical form:



Step 1: Rewrite the rational expression in radical form.Step 2: Simplify: 8 to the first power equals 8.Step 3: Factor 8 into (2)(2)(2).Step 4: Since there are three factors of 2 under the cube root, the radical expression will simplify to 2.

Answer: 2

Simplifying using exponent laws:

Step 1: Rewrite the expression and factor 8 into (2)(2)(2).Step 2: Write (2)(2)(2) in exponential form.Step 3: Multiply 3 and 1/3 (remember that when an exponent is taken to another power, the two powers are multiplied) to get 1. Simplify 2 to the first power to 2.

Answer: 2


Step 1: Rewrite the expression as a radical expression.Step 2: Simplify 322 : 32 32 = 1,024.Step 3: Factor 1,024 into (4)(4)(4)(4)(4). Always try to factor the number under the radical into the same number of factors as the root.Step 4: Since this is a 5th root and there are five 4s under the radical, the radical expression will simplify to 4.


Answer: 4

Activities that can help reinforce the concept of rational exponents are as follows.

1. Take 30 index cards, and make three piles of ten cards each. Write expressions with rational exponents on the first set of ten cards. On the second set, write the radical form for each expression you wrote on the first ten cards. On the third set, write the simplified answer to the ten expressions. Shuffle all of the cards and lay them face down on a table. Each player takes turns flipping three cards. If all three cards go together, the player has made a match and gets to try again. If not, it is the next player's turn.

2. Name expressions with rational roots, and ask the student to factor the base into the same number of identical factors as the root. For example, if the root is 1/6, the student should factor the base into 6 identical factors, so that when the factors are rewritten in exponential form, and the exponent is multiplied by the root, the product is 1.

Polynomials: SubtractionA monomial is the product of a number and an unknown variable or unknown variables. 6xy is a monomial. The sum or difference of two or more monomials is called a polynomial.

Here is an example of a polynomial: Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are monomials that have the same variable or variables for which the variable or variables have the same exponent.

To subtract polynomials, first write the polynomials as one long polynomial. Then distribute the subtraction sign through the second polynomial. Finally, combine like terms. Practice by subtracting the following polynomials.

Example 1:

Step 1: Set up the two polynomials as one long polynomial. Since the problem is to subtract one polynomial from another, the second polynomial in the problem must be written first.Step 2: Distribute the subtraction sign through the second polynomial. This involves changing the sign of each term in the second polynomial.Step 3: Rewrite the polynomial after changing the signs in the second polynomial.


Step 4: Combine like terms.

Answer: 5p + 9

Example 2: Subtract four times a number decreased by ten from eight times the same number less six.

Step 1: "Four times a number decreased by ten" can be written (4x - 10).Step 2: "Eight times the same number less six" can be written (8x - 6).Step 3: Now the problem reads: Subtract (4x - 10) from (8x - 6).

Step 4: Set up the polynomials as one long polynomial.Step 5: Distribute the subtraction sign through the second polynomial. This involves changing the sign of each term in the second polynomial.Step 6: Rewrite the entire polynomial after changing the signs in the second polynomial.Step 7: Combine like terms.

(4x - 10) subtracted from (8x - 6) equals 4x + 4.

Answer: 4x + 4 Example 3: Find area of the shaded region.

Step 1: Determine the area of the large rectangle by multiplying the length (2x + 7) by the width (3x). This involves multiplying each term in (2x + 7) by 3x.Step 2: Determine the area of the small rectangle by multiplying the length (2x - 2) by the width (x). This involves multiplying each term in (2x - 2) by x.


Step 3: Now subtract the area of the small rectangle from the area of the large rectangle. Remember to put the second polynomial in parentheses since this is subtraction.Step 4: Distribute the subtraction sign through the second polynomial. This involves changing the sign of each term in the second polynomial.Step 5: Combine like terms.

Answer:

Equations: Order of OperationsAn equation is a statement in which two numbers or two expressions are set equal to each other. For example,5 + 3 = 8 and 16 = 3c + 4 are equations.

When solving equations, find the value of the variable by getting the variable alone on one side of the equal sign. To do this, undo any operations on the variable by using the inverse operation. Any operation done on one side of the equal sign must be done on the other side of the equal sign in order to keep the statement true.If a number has been added to the variable, subtract the number from both sides of the equation.

m + 3 = 5 - 3 - 3 m = 2

If a number has been subtracted from the variable, add the number to both sides of the equation.b - 7 = 9 + 7 + 7 b = 16

If a variable has been multiplied by a nonzero number, divide both sides by the number.

If a variable has been divided by a number, multiply both sides by the number.

When solving 2-step equations, we must first undo the addition or subtraction using the inverse operation, then undo the multiplication or division:


Example 1: Solve the equation for t. 5(t - 4) = t + 12

Step 1: Multiply 5 times the terms inside the parenthesis.

Step 2: Add 20 to both sides of the equation

Step 3: Subtract t from both sides of the equation

Step 4: Divide both sides of the equation by 4.

Answer: t = 8

Example 2: Evaluate the expression for c = 3:2(c + 4) + 2(15)

(1) 2(3 + 4) + 2(15)(2) 2(7) + 2(15)(3) 14 + 30(4) 44

Step 1: Substitute 3 in place of 'c' in the expression.Step 2: Add the numbers in parentheses.Step 3: Rewrite the equation after performing all multiplications in order from left to right.Step 4: Add 14 and 30 to get 44.

Answer: 44


Functions: LinearThis skill requires students to solve real world problems involving two linear functions. A linear function is a function whose graph forms a non-vertical straight line.

Story problems are often very difficult for students to master. It may be beneficial first to confirm that the student is comfortable with solving two linear equations outside of a real world context. To do this, students must review how to solve linear equations. In some variable equations, students are given the value of one variable. Letters expressing unknown values in equations are called variables. To find the value of the second variable, substitution must be used.

If y = 4x, and x = 9, substitute the given value of x in the equation, y = 4x. The result is y = 4(9). Calculate the right side of the equation to get y = 36. Now the values of both variables are known: x = 9 and y = 36.

To check the answer, set up the equation y = 4x as if x was an unknown variable: 36 = 4x. Divide each side of the equation by 4. The result is x = 9, so the values for x and y are both correct.

Here is another situation where the value of a variable is substituted into an equation to solve for x.

Example 1:

Solve for x.

x = 3y + 9y = - 2

Solution: Substitute - 2 in place of y in the first equation. Then, solve for x.x = 3(- 2) + 9x = - 6 + 9x = 3

Answer: x = 3

This concept can be used to solve two equations that include the same two variables.

Example 2:

Solve for x and y in the following equations.y = 2x - 33x = y + 10

Solution:If y = 2x - 3 and 3x = y + 10, there is not a number to substitute in for either of the variables. So, an expression must be substituted for a variable. In this case, the y variable is already isolated. It is equal to 2x - 3. This can be substituted in for y in the second equation.

(1) 3x = y + 10SD Algbra-Questions, Answer Key, Study Guide of 104

3x = (2x - 3) + 10(2) 3x = 2x + (- 3 + 10) 3x = 2x + 7(3) 3x = 2x + 7 -2 x -2 x x = 7(4) y = 2x - 3 y = 2(7) - 3(5) y = 14 - 3 y = 11

Step 1: Substitute 2x - 3 for y in the second equation.Step 2: Simplify the equation by removing the parentheses and combining the like terms - 3 and 10.Step 3: Isolate the variable by subtracting 2x from each side and combining like terms.Step 4: Since there is now a value for one of the variables, it can be substituted back into either of the original equations. In this example, the first equation was used.Step 5: Calculate the right side of the equation to get y = 11.

Answer: x = 7 and y = 11

Several questions from this skill relate to measurement concepts. The formula for perimeter is occasionally needed to find solutions to story problems. Perimeter is the measurement around a figure.

Example 3:

Find the perimeter for the figure below.

Solution:To calculate the perimeter of a figure, add the lengths of all the sides of the figure. For example, this figure has four sides measuring 3 inches, 7 inches, 3 inches, and 7 inches.

P = 7 + 3 + 7 + 3 = 20

Answer: The perimeter of this figure is 20 inches.

Story problems that involve perimeter and two linear equations can be solved by writing two equations, then using the substitution method that was reviewed earlier.

Example 4:Emilio built a square sandbox for his dog with a perimeter of 15 m. Since his dog has grown, he wants to rebuild the sandbox so that each side is 2 m longer than the original sandbox. Before he buys the materials to build the new sandbox, he needs to know the perimeter. What is the perimeter of the new sandbox?

Solution:SD Algbra-Questions, Answer Key, Study Guide of 104

To solve this problem it is probably best to start with drawing a picture. Because we do not know the length of the sides of the old enclosure, they can be assigned a variable, x. The new enclosure must have sides that are 2 m longer than the original, so the sides can be written as x + 2.

(1) For the old enclosure, x + x + x + x = 15, or 4x = 15 For the new enclosure, (x + 2) + (x + 2) + (x + 2) + (x + 2) = y, or 4(x + 2) = y.(2) 4x = 15, so x = 3.75(3) 4(3.75 + 2) = y(4) y = 23

Step 1: An equation can be written for the perimeter of the new and old enclosures based on the information in the problem. For the new enclosure, we do not know the perimeter so we give it a different variable, y.Step 2: The x must be isolated in the first equation so that it can be substituted into the second.Step 3: This value can be placed in the second equation for x.Step 4: Calculate.The perimeter of the new enclosure will be 23 m.

Answer: 23 m

Another common application of finding the solution to two equations in solving real world problems involves money.

Example 5: Ariel wants to go snowboarding this weekend. Ice Mountain charges $24 for a lift ticket and $8 per hour of snowboarding. Flurry Ridge charges $29 for a lift ticket and $6 per hour of snowboarding. Ariel must get a lift ticket to snowboard and she wants to snowboard for 5 hours. Which ski area would be cheaper for Ariel's snowboarding trip?

(1) For Ice Mountain, the cost = 24 + 8h. For Flurry Ridge, the cost = 29 + 6h.(2) For Ice Mountain, 24 + 8(5) = 64 For Flurry Ridge, 29 + 6(5) = 59.

Step 1: Set up two equations, one for each snowboarding location. We can use the variable h for the number of hours.Step 2: Since Ariel wants to snowboard for 5 hours, we can substitute 5 in for h in each of the equations and solve.

Answer: Ice Mountain will cost her $64 and Flurry Ridge costs her only $59, so Flurry Ridge is cheaper.

To help students understand the application of linear equations in solving real world problems, students SD Algbra-Questions, Answer Key, Study Guide of 104

can find the costs per unit of time or distance of different services with the help of a telephone book, the newspaper, or the internet. If specific cost information is not listed, call the company for information. For example, a car rental company may charge a base fee of $35 per day and $0.59 per mile or a catering company may charge a rental fee of $100 for materials and then $60 per hour for waitstaff services. Then, have students calculate which would be the cheapest and which would be the most expensive for a specified distance a rental car may drive or amount of time a catered party may last.

Quadratic Equations: Real World ProblemsA quadratic equation is a second-order polynomial equation in a single variable. Second-order means that one of the terms contains the variable to the power of two and no other exponents are greater than two. Students will apply quadratic equations to evaluate real world problems in this question set.

Real world problems are often difficult for students to master. It may be beneficial first to confirm that the student is comfortable with quadratic equations outside of a real world context. To do this, students must first review the basic principles of quadratic equations, including how to distribute terms in order to obtain the equation in the proper form, how to factor, and eventually, how to solve for the variable.

Review of Definitions:

• A variable is an unknown number in an equation and is usually represented by a letter such as x.• A coefficient is a number representing a fixed, or constant, value that is multiplied with a variable.• A polynomial is a term or sum of terms. Each term is either a number or the product of a number and one or more variables. For example, 2x2 + 3 is a polynomial with two terms, 2x2 and 3.• A binomial is a polynomial with two terms, and a trinomial is a polynomial with three terms.• Factoring is the breaking up of quantities into products of their component factors. One way to think of factoring is as the opposite, or inverse, of multiplying.

Quadratic functions are written in the form:

where f(x) is read "f of x." The letters a, b, and c are constants and a 0.When solving quadratic equations, replace the y or f(x) with zero. It is important to remember that all quadratic equations have two answers. However, the answers could be imaginary (contain a square root of a negative number), positive, or negative. When dealing with real world problems, the positive answer is typically the only one that would make sense in the context and would therefore be the only correct answer.

Below are a few examples of quadratic equations:

Sometimes, quadratic equations come in a form where they have already been factored. Some examples of these equations are shown below.


(x + 5)(x - 1) = 7or

2x(x + 6) = 54

In order to find the solution to quadratic equations, they have to be set equal to zero. To get the equations in this form, multiply out the terms to eliminate the parentheses and subtract appropriately until the polynomial is equal to zero. Once an equation is written in the proper form, the next step towards solving for the variable is factoring the equation.

Example 1:Factor completely and solve for y.

y2 + 4y - 2 = 3

Step 1: Subtract 3 from both sides of the equation so that the equation is equal to zero.Step 2: Determine the factors. Since the coefficient of the y2 term is equal to 1, focus on the last term, in this case, - 5. If factors of - 5 can be found that add up to the coefficient of the "y" term (in this example, 4), the trinomial can be factored. The factors of - 5 are 5 and - 1 or - 5 and 1. When we add the first pair of numbers, 5 and - 1, we get 4, the number we are seeking.Step 3: Check your answer. To check the result, multiply the binomial factors together. Binomials are multiplied using the FOIL method. FOIL is the abbreviation for First, Outside, Inside, and Last.

• Multiply the First terms of each binomial (y y = y2 )• Multiply the Outside terms of each binomial (y -1 = -1y)• Multiply the Inside terms of each binomial (5 y = 5y)• Multiply the Last terms of each binomial (5 - 1 = - 5)• Combine like terms to simplify the polynomial.

Step 4: Set each factor equal to zero. When we multiply any value by zero the result is zero. Therefore, if either factor is equal to zero, the entire expression is equal to zero. Step 5: Solve each of these equations by isolating the variables.

Answer: y = - 5 and y = 1

A special type of polynomial expresses the difference of two perfect squares. Polynomials of this type are easily factored once the pattern is remembered.


Since each term in the polynomial is a perfect square, the square root of each term (in this case x and 6, respectively) will be used in the following way. The original polynomial is factored as (x + 6)(x - 6). Notice that if these terms are multiplied together using the appropriate FOIL method, the original polynomial is formed. Polynomials that are in the "difference of squares" form, that is, a2 - b2 , can always be factored as the sum of the square roots, (a + b), multiplied by the difference of the square roots, (a - b).

Using Quadratic Equations in Real World Problems:

When solving real world problems, clues from the text of the problem must be used to set up the equations. Quadratic equations are often used to find solutions to story problems. The story problems in this set also occasionally require the use of formulas for the area of a rectangle, the area of a triangle, or the Pythagorean Theorem.

Area is the measure of the interior of a two-dimensional region and is calculated in square units. The formulas used to find the area of a rectangle and a triangle are shown below.

Area of a rectangle = length x width, or A = lw. A square is a rectangle with four equal sides, so the formula for the area of a square is often shown as Area of a square = side2 , or A = s2 because the length is equal to the width for all squares.

Area of a triangle =1/2 x base x height, or A = (1/2)bh. The height of a triangle is the length of the line segment that extends from the vertex opposite the base and is perpendicular to the base, as shown below.

The Pythagorean theorem is used to find the lengths of the sides and hypotenuse of a right triangle. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs of that triangle. A right triangle is a triangle containing a 90º angle, the hypotenuse is the side opposite the right angle (longest side of the triangle), and the legs are the two other sides.

Using the diagram above, it is important to remember that a and b always denote legs of a right triangle and c always denotes the hypotenuse of a right triangle.


Sometimes, the length or width are unknown in an area problem or the leg is unknown in a right triangle problem. Examples of these types of problems are provided below.

Example 2:Colleen is swimming in the ocean. She swam 4 miles south and then turned and swam 3 miles west. If she turns and swims straight back to the point where she started swimming, how many more miles does she need to swim?

Step 1: Determine the values of a, b, and c. In this case, a and b are known and c is unknown.Step 2: Substitute the values of a and b into the Pythagorean theorem.Step 3: Square 3 (3 3 = 9) and square 4 (4 4 = 16).Step 4: Add 9 and 16 to get 25.Step 5: Take the square root of each side of the equation. The square root of 25 is 5 and the square root of c2 is c.

Answer: She will swim 5 miles.

Example 3:Find the height of a triangular sail that has an area of 60 m2 and a base of 8 m.

Step 1: Begin with the formula for the area of a triangle.Step 2: Substitute in 60 for the area and 8 for the base.Step 3: Simplify the equation by multiplying 1/2 and 8 to get 4.


Step 4: Isolate the height by dividing each side of the equation by 4.

Answer: The height of the sail is 15 m.

When working with story problems, use a variable for the unknown value and clues within the wording to write the equations that will need to be solved. For example, if a problem reads "the length is 4 m longer than the width," then the length can be represented by w + 4. After an entire equation is formed, it can then be solved.

Example 4:The width of a rectangular cake pan is 3 inches shorter than the length of the pan. The area of the cake pan is 108 in.2 What is the length of the cake pan?

Solution:

(1) (2) x(x - 3) = 108(3) (x)(x) - (x)(3) = 108 x2 - 3x = 108(4) x2 - 3x - 108 = 0(5) (x - 12)(x + 9) = 0(6) x - 12 = 0 and x + 9 = 0(7) x = 12 and x = - 9

Step 1: Drawing a picture of the cake pan with the information in the problem may help in solving the problem. The length is the unknown value, so it can be represented by the variable x. The width is 3 inches shorter than the length, so it can be written as x - 3.Step 2: Since this problem deals with the area, length, and width of a rectangle, we must use the formula for the area of a rectangle, Area = length x width, and substitute in x for the length, x - 3 for the width, and 108 in.2 for the area.Step 3: Distribute the x and simplify.Step 4: Subtract 108 from both sides to put it in the proper form.Step 5: Factor the equation.Step 6: Set each polynomial equal to zero.Step 7: Isolate the variable in each problem.

Answer: The answer x = - 9 does not make sense because the length of a cake pan cannot be a negative value. That answer is discarded and the correct answer must be x = 12. The length of the cake pan is 12 in.

To help students understand the application of quadratic equations in solving real world problems, help the student find the missing dimensions of various objects around the house when given the area and information about one of the dimensions. For example: Measure a picture frame, door, etc., calculate the area of the object, and create a relationship between the dimensions (the length of the door is twice the width). Then have the student find the missing dimension when given the area and the relationship between the dimensions. An activity to see how the Pythagorean theorem works is to lean a board or ladder against a wall to create a right triangle. Measure two dimensions of the triangle and use the Pythagorean theorem to determine the length of the third side of the triangle.


Functions/Relations - AA relation is a set of ordered pairs that represent a relationship between the elements of the two sets. A function is a special type of relation, where each element of the first set (x-values) corresponds to an unique element of the second set (y-values). The first set of numbers is commonly known as the input and the second set as the output. The input, or x-values, are entered into the equation. Once evaluated, the result is the output, or y-values. In other words, in order for a relation to be a function, for each x-value there can be no more than one value of y. Some examples of relations are given below, with input values in A mapped to output values in B.

Relations 1 and 2 are functions, while relation 3 is not a function. The input value - 1 in relation 3 is matched to more than one output value (3 and 5), so the relation is not a function.

Example 1:Which of the following relations is not a function?

Solution:If there is a value of x resulting in more than one value of y, the relation is not a function. This only occurs in the third set of numbers with (9, - 1) and (9, 4). Therefore, set C is not a function.

Answer: Set C is not a function.

Example 2:Which of the following points, if removed from the set, would make the set a function?

Solution:The ordered pairs (- 4, 5) and (- 4, 4) have the same x values but different y values. Therefore, if either point is removed from the set, the remaining ordered pairs will represent a function.

Answer: Remove either (- 4, 5) or (- 4, 4).


Equations With Two VariablesThis study guide will focus on solving equations that contain two variables.

Remember:Variables are letters or symbols that represent numbers that are unknown.Expressions are variables or combinations of variables, numbers, and symbols that represent a mathematical relationship. Expressions do not have equal signs, but can be evaluated or simplified.

Example: y - 9Equations are expressions that contain equal signs. They can be solved, but not evaluated.

Example: 5 + x = 13Like terms are those terms that have the same variable(s) in common, or no variable. Like terms can be combined.

Examples: 2 and 3 are like terms, 2x and 3x are like terms, but 2 and 3x are not like terms.

Solving EquationsTo solve an equation, it is necessary to "undo" what was done to the variable in question. Another way to think about this is to do the order of operations in reverse.

P = Parenthesis, E = Exponents, M = Multiplication, D = Division, A = Addition, S = Subtraction.**In the normal direction, first evaluate multiplication and division from left to right (whichever comes first), then evaluate addition and subtraction from left to right (whichever comes first).**To "undo" what was done, reverse the direction.

•Addition "undoes" subtraction.•Subtraction "undoes" addition.•Multiplication "undoes" division.•Division "undoes" multiplication.

Addition/Subtraction Properties - Adding or subtracting the same real number to each side of an equation will result in an equivalent equation.

Example 1: Solve 16 = 5 + x - 8 for x.

Step 1: Begin to isolate the variable, x, by adding 8 to both sides of the equation. Remember to only add and subtract like terms, so the 8 must be added to the - 8 and the 16.Step 2: Subtract 5 from both sides of the equation to completely isolate the variable, x.

Answer: x = 19


Multiplication/Division Properties - Multiplying or dividing each side of an equation by the same (nonzero) number will result in an equivalent equation.

Example 2: Solve the following equation for t.

Step 1: Multiply both sides of the equation by 17 to eliminate the fraction.Step 2: Divide both sides of the equation by 4 to isolate the variable, t.

Answer: t =136.

Solving Equations With Two VariablesIn order to solve equations that contain two variables, the student will need to solve for one variable in terms of the other. This means that the answer may not be entirely numeric.The same properties as described above should be used when solving for a given variable in a two-variable equation. Students should treat the second variable as if it were a number.

Example 3: Solve 30w - 9v = 6 for w.

Step 1: Begin to isolate w by adding 9v to both sides of the equation. Remember to only combine like terms.Step 2: Completely isolate the variable, w, by dividing by 30 on both sides of the equation. Step 3: Simplify the solution. Since 6, 9, and 30 can all be divided by 3, divide the numerator and denominator of the fraction by 3.

Example 4: Solve the following equation for r.


Step 1: Begin to isolate r by adding 7 to both sides of the equation. Remember to only combine like terms. 10s + 7 cannot be combined because they are not like terms. Step 2: Completely isolate r by multiplying both sides of the equation by 7.Answer: r = 7(10s +7)

Example 5: Solve the following equation for a.5(a - b) = 14b

Step 1: Questions of this type involve using the distributive property. The distributive property states that for all numbers a, b, and c, a(b + c) = ab + ac. Therefore, begin by multiplying 5 by a and by b. Step 2: Next, begin to isolate a by adding 5b to both sides of the equation. 14b and 5b are considered like terms so they can be added together to get 19b.Step 3: Completely isolate the variable, a, by dividing both sides of the equation by 5.

Example 6: Solve the following equation for x.8xy - 9x = 20

Step 1: Questions of this type involve using the distributive property to write x as a factor. Begin by factoring out the x as a common factor.Step 2: Isolate the variable, x, by dividing both sides of the equation by (8y - 9).

Literal EquationsThis study guide will focus on finding solutions to literal equations. Students will be given an equation, usually a formula, and be asked to solve the equation for a particular variable. This skill is especially useful when students need to find missing values in science experiments and technical fields.

Literal equations are comprised of literals or multiple letters, also called variables. These literals can represent known constants or numbers, or be extra variables, letters that represent a range of values to be determined. So a literal equation uses a bunch of letters to represent a specific relation. They are usually referred to as formulas.

It may be beneficial to confirm that the student is comfortable with solving linear equations, equations containing one or two variables that do not contain exponents, before beginning to solve literal


equations. A quick review is provided below.

Solving Linear Equations

To solve a linear equation, it is necessary to "undo" what was done to the variable in question. Another way to think about this is to do the order of operations in reverse.

P = Parenthesis, E = Exponents, M = Multiplication, D = Division, A = Addition, S = Subtraction.**In the normal direction, evaluate multiplication and division from left to right (whichever comes first). Evaluate addition and subtraction from left to right (whichever comes first).**To "undo" what was done, reverse the direction.

•Addition "undoes" subtraction.•Subtraction "undoes" addition.•Multiplication "undoes" division.•Division "undoes" multiplication.

Example 1:Solve for x.x + 5 = 7

Step 1: It is necessary to get x by itself on one side of the equation and everything else on the other side. Undo the "+ 5" by subtracting 5 from both sides of the equation.Step 2: x is now isolated and the other side of the equal sign is simplified.

Answer: x = 2

Solving Literal Equations

The methods used for solving literal equations are the same as those used in solving linear equations, except fewer simplifications can be made since the variables cannot be combined. The goal is to solve the formula for a given variable in terms of the other variables.

Example 2:


Step 1: Square both sides to eliminate the radical.Step 3: Take

the square root of both sides to eliminate the squared term on the side of the equation with the x-terms.Step 5: Divide both

sides by -1 to eliminate the negative.

Example 3: Solve for w.

P = 2w + 2l.

Step 1: To isolate the w, subtract 2l from both sides of the equation to get P - 2l = 2w.Step 2: Then divide both sides of the equation by 2.

A good way to practice this skill is to look up formulas used in different fields (physics, electronics,


chemistry, etc.) and practice solving each one for a different variable. Some formulas may be more complicated than others, so try to stick to those that have the four basic operations, addition, subtraction, multiplication, and division.

Missing Elements - DEquations are presented with either a number or operational symbol omitted. Students must solve the equation by determining what is missing.

It may be useful to develop a series of mathematical equations with either a number or operational symbol missing. Help the student replace the missing elements.

Example 1: 6 ? 4 = 10

The missing element is the plus symbol (+), we know this because 6 + 4 = 10.

Example 2: Find the operational symbol from the multiple choices below.99 ? 3 = 33

A. +B. -C. xD. ÷

Answer: D, since 99 ÷ 3 = 33.

Example 3: Find the missing number.

4 x (5 + 2) = 20 + ?

(1) 5 + 2 = 7(2) 4 x 7 = 20 + ?(3) 4 x 7 = 28(4) 28 = 20 + ?

Step 1: First determine the value within the parentheses.Step 2: Rewrite the equation with the new value in place of the parentheses.Step 3: Perform the operations on the opposite side of the value we are finding. 4 x 7 = 28Step 4: Rewrite the equation with the new value in place of 4 x 7.

Answer: 8

Example 4: Find the missing number.5 x (? + 3) = (2 x 5) + (3 x 5)

(1) 2 x 5 = 10 and 3 x 5 = 15(2) 5 x (? + 3) = 10 + 15(3) 5 x (? + 3) = 25(4) 5 x (2 + 3) = 25


Step 1: Perform operations within the parentheses.Step 2: Rewrite the equation with the new values in place of the parentheses.Step 3: Add 10 and 15 to determine the value on one side of the equation.Step 4: We can now find the missing value. We know that 5 x 5 = 25, so we know the value within parentheses must equal 5. So, since 2 + 3 = 5, the missing value is 2.

Answer: 2

Radicals and RootsMastering roots and radicals is an essential step toward learning advanced mathematics concepts. A radical sign looks like a check mark with a line attached to the top. The radical sign is used to communicate square roots.

The following rules are required to perform operations with roots and radicals.

1. If x multiplied by x equals y, then x is a square root of y. For example, 6 multiplied by 6 is 36, so 6 is a square root of 36. In fact, 36 is called a perfect square because its square root, 6, is a whole number. Most algebra text books contain a table of perfect squares.

2. -3 and 3 are both square roots of 9 because -3 x -3 = 9 and 3 x 3 = 9. 3 is referred to as the principal square root because it is the positive square root of 9.

3. To find the simplest radical form of a radical expression, factor the number under the radical sign (the radicand). The square root of 45 could be factored to be the square root of 9 multiplied by the square root of 5. The square root of 9 multiplied by the square root of 5 can be simplified further by finding the square root of 9. The result is 3 (the square root of 9) multiplied by the square root of 5.

Example 1: Find the equivalent form.

Solution: Multiply the numbers under the radical symbols.

Example 2: What symbol would best replace the ? in the given statement?

A. =B. <C. >

There are two methods that can be used to solve this problem. Each method is shown and explained below.


Solution Method 1:

Step 1: Simplify the square root of 12 by making it the square root of 6 x 2.Step 2: Further simplify the square root of 6 x 2 by making it the square root of 3 x 2 x 2. (If you multiply 3 x 2 x 2, you will get 12.)Step 3: The square root of 3 x 2 x 2 becomes 2 times the square root of 3, because the square root of 2 x 2 is 2 and the 3 must remain under the square root symbol.Step 4: Two times the square root of 3 can also be written as the square root of 3 plus the square root of 3.Step 5: Now, we can make a comparison. We know that the larger a number is, the larger that number's square root will be. We can determine that the square root of 5 plus the square root of 7 will be greater than the square root of 3 plus the square root of 3 because 5 and 7 are both larger than 3.

Answer: C

Solution Method 2:

Step 1: Estimate the square root of 5, the square root of 7, and the square root of 12. This estimation can be done using a calculator.Step 2: Add together the 2.24 and the 2.65 to get 4.89.Step 3: Replace the question mark with the > symbol because 4.89 is greater than 3.46.

Answer: C

Example 3: Solve for the value of x.


Step 1: The square root of 36 is 6 because 6 x 6 = 36.Step 2: Subtract 6 from each side of the equation to isolate the square root of x.Step 3: The square root of x is equal to 8. We can replace the x with 64, since the square root of 64 is 8.Step 4: Since the square root of 64 equals 8, the value of x is 64.

Answer: x = 64

Ratio/Proportion - CA ratio is a comparison of two numbers expressed as a quotient. Ratios can be written in three ways: a fraction (3/5), a ratio (3:5), or a phrase (3 to 5). Like fractions, ratios refer to a specific comparison. The ratios 3/5, 3:5, and 3 to 5 (as in "the ratio of cellos to violins was 3 to 5") all express the same ratio or comparison. A proportion reflects the equivalency of two ratios. The ratio 3/5 expresses the same proportion as the ratio 15/25.

To understand how ratios operate, students need to understand equivalent fractions. Fractions represent portions or parts. For every fraction, there is a corresponding portion. The fraction 1/2 communicates one portion out of two, but this specific portion can also be communicated by the fractions 2/4, 3/6, 8/16, 10/20, etc. All of these fractions are equal to 1/2 because the relationship between the numerator and denominator in 1/2 is the same relationship between the numerators and denominators in 2/4, 3/6, 8/16, and 10/20. Ratios and proportions operate in a similar manner. The ratio 2:5 communicates a specific portion. The ratio 4:10 communicates the same portion because 4:10 reduces to 2:5.

Example 1: Sandra has 15 lollipops and 25 jellybeans. What is the ratio of lollipops to jellybeans?

Answer: There are 15 lollipops to 25 jellybeans, so the ratio of lollipops to jellybeans is 15:25.

Proportions occur when two ratios are equal. In a proportion the cross products of the terms are equal.Example 2: Is the following proportion True or False? 1/3 = 3/9

The cross products are both equal to 9, so the proportion is TRUE. If the cross products were not equal, the proportion would be FALSE.

Sometimes you must find the value of a variable in a proportion. To solve the proportion, you must find the value of the variable that makes both ratios equal.

Example 3: What is the value of a?9/12 = a/48


Step 1: Find the cross products. Multiply 48 by 9 and 12 by 'a'.Step 2: 48 x 9 = 432 and 12 x a = 12a. Rewrite the equation with the new products.Step 3: Divide each side of the equation by 12 to isolate the variable 'a'.Step 4: Divide 432 by 12 to get a = 36.

Answer: a = 36

Example 4: Apples are 13 for $3.35. What will 32 apples cost? (Round answer to the nearest hundredth).

Step 1: Write the appropriate proportion. Let b represent the cost of 32 apples.Step 2: Write the cross products. Multiply b by 13 and multiply 3.35 by 32.Step 3: Rewrite the equation with the new values.Step 4: Divide each side of the equation by 13 to isolate the b.Step 5: 107.20 ÷ 13 = 8.246153846 We only need to write 3 decimal places because we are going to round to the hundredths place (the second decimal place). The three dots above represent that the decimal number continues on past that point.Step 6: Round 8.246 to the hundredth place. The 6 tells us to round the 4 up to a 5. 8.246 approximately equals 8.25.

Answer: $8.25

Take a look at the solution. Does it make sense that 32 apples would cost $8.25? We know that 32 apples would cost more than twice as much as 13 apples (13 x 2 = 26), but less that three times as much (13 x 3 = 39). The solution is more than twice as much as $3.35 (2 x $3.35 = $6.70), but less than three times as much (3 x $3.35 = 10.05), so our answer makes sense.

Example 5: Rachel scored 25% of the team points. If the team scored 40 points, how many did Rachel score?


Step 1: Write a proportion to represent the problem.Step 2: Cross multiply. This involves multiplying the bottom of one ratio by the top of the other and vice versa. Perform the multiplication.Step 3: Divide both sides of the equation by 100.

Answer: 10 points

Function/Pattern - CStudents must identify patterns in numbers (such as missing elements) and shapes, and then draw conclusions based upon the identified pattern.

It may be helpful to develop a series of number patterns and have the student identify the pattern and determine the correct equation. For example, the pattern

5, 7, 9, 11This sequence follows a pattern of adding 2 to an odd number.

3 + 2 = 55 + 2 = 77 + 2 = 99 + 2 = 11

To practice critical thinking skills, have the student not only find the patterns, but also analyze the patterns. For instance, have him or her identify the patterns of 3 lists of numbers. Then, ask the student which pattern will need the fewest additional numbers to reach a specific number. For example, consider the following 3 patterns. Which pattern will need the fewest additional numbers to reach 95?

(a) 71, 74, 78, 81, 85, ...(b) 2, 5, 11, 23, 47, ...(c) 5, 10, 15, 20, 25, ...

For (a), the pattern is to add three and then to add four. It will take 3 more numbers to reach 95. For (b), the pattern is to double the number and then add 1. It will take 1 number to reach 95. For (c), the pattern is to add five. It will take 14 numbers to reach 95. Therefore, (a) would be the correct answer.

Example: Which of the following equations can be used to find the numbers in this table?

A. B = A + 9SD Algbra-Questions, Answer Key, Study Guide of 104

B. B = 2A + 1C. B = 3A - 6D. B = A + 11

Solution: To get from 7 to 15, you must either add 8 or multiply 7 by 2 then add one. To get from 8 to 17, you must either add 9 or multiply 8 by 2 then add one. Since the pattern "multiply by 2 then add one" is common in both of the first two rows, check to see if this pattern works for the rest of the rows in the table. It does work for the rest of the table, so the answer is B.

Inequalities - AAn inequality is a number sentence that uses "is greater than", "is less than", or "is not equal to" symbols. For example, 6n > 4 is a number sentence with an inequality symbol.

It may be useful to review the inequality symbols.

Example 1: Solve for y: 8y > 40

Get the variable being solved for (y) on one side of the inequality and the whole number on the other. To do this, divide both sides by 8.

The correct answer is that y is greater than 5.

Inequalities can be represented as a value on a number line. The following number line represents the inequality

Example 2: Which inequality represents the value shown on the number line below?

A. n < 3B. n > 3C. n = 3

The answer is A. n < 3 because the dot on the number line is open.


Functions: QuadraticA quadratic function is a function that contains polynomial expressions for which the highest power of the unknown variable is two.

Quadratic functions are written in the form:

or

f(x) is read "f of x."

Here are a few examples of quadratic functions:

Since the value of f(x) (or y) depends on the value of x, the dependent variable is f(x) and the independent variable is x.

Domain and Range:

The domain of a function is the set of values that can be substituted for the independent variable (x) of a function. When dealing with the coordinate pairs of a function, the domain is the set of all x values of the function.

Example 1: Find the domain of the following function.

Since the domain deals with all of the x values of a function, we need to determine all of the x values of this function. The domain of the function is:

The domain of a function is generally written in numerical order from smallest to largest.

Example 2: Find the domain of

Since any real number that is substituted in place of the independent variable (x) will result in another real number, the domain of this function is: D = The set of all real numbers.

The range of a function is the set of values that can result from the substitutions for the independent variable. When dealing with the coordinate pairs of a function, the range is the set of all y values of the function.

Example 3: Find the range of the following function.


Since the range deals with all of the y-values of a function, we need to determine all of the y values of this function. The range of the function is The range of a function is generally written in numerical order from smallest to largest. If a y-value is repeated in an ordered pair, it is only listed once in the range.

Example 4: Find the range of Since any real number that is substituted in place of the independent variable (x) will result in another real number (g(x)), the range of this function is: R = The set of all real numbers.

Evaluating Quadratic Functions:

In mathematics, evaluate means to substitute values in for the variables and calculate a result. So, to evaluate a quadratic function, we need to substitute a value in for the independent variable (x) and calculate the result.

Example 5: Evaluate the quadratic function at f(3).

Step 1: The directions told us to evaluate the quadratic function at f(3), so we substitute 3 in place of all of the x's in the quadratic function.Step 2: Following the order of operations, we perform operations on exponents first, then we multiply.Step 3: Once again, we follow the order of operations which states that multiplying comes before adding and subtracting.Step 4: The final step in the order of operations states that adding and subtracting are completed by working from the left to the right. First we will add 18 and 9. Then we subtract 8 from that sum. We now know that f(3) = 19.

It is possible to evaluate a function and still have a variable in the answer.

Example 6:

Step 1: Substitute (x + 4) in place of all of the x's in the quadratic function.Step 2: Applying the order of operations, we must do all operations inside the parentheses. So we add 4 and 3.Step 3: To take (x + 7) to the second power, we must multiply (x + 7) by itself.Step 4: In order to multiply (x + 7) by itself, we use the FOIL method. Multiply the "first" terms together(x times x). Then multiply the "outer" terms (x times 7). Next, multiply the "inner" terms (7 times x). Finally, multiply the "last" terms (7 times 7).


Step 5: Since 7x and 7x are like terms, we can add them together to get 14x. Also, 49 and 5 are like terms, so we can add them together to get 54.

Compositions of Quadratic Functions:

The composition of functions is the operation of first applying one function, then applying the other. Compositions of functions are denoted in two ways. One of those ways is which can be read, "the composite of g and f of x." The other way is g(f(x)), which can be read, "g of f of x."

Example 7:

Step 1: Since we are trying to find "g of f of x," we can substitute in place of f(x) in g(f(x)) to get Then, we can substitute that term in place of x in the g(x) = 6x function.Step 2: Use the distributive property to multiply each term in the parentheses by 6.

Example 8: Find f(g(x)) if f(x) = 2x + 6 and g(x) = 7x - 2.

(1) f(7x - 2) = 2(7x - 2) + 6(2) f(7x - 2) = 14x - 4 + 6(3) f(g(x)) = 14x + 2

Step 1: Substitute (7x - 2) in place of g(x) to get f(g(x)) = f(7x - 2). Then substitute (7x - 2) in place of all x's in the f(x) = 2x + 6 function.Step 2: Use the distributive property to multiply 7x and -2 by 2.Step 3: Since -4 and 6 are like terms, they can be added together to make 2. So, f(g(x)) = 14x + 2.

Compositions of functions can be evaluated if we are given the value of the variable.

Example 9:

Step 1: To determine the value of f(g(4)), we must first evaluate g(4). We substitute 4 in place of x inStep 2: Following the order of operations we square 4 (4 x 4 =16) and add one to the

answer. g(4) = 17.Step 3: Now we substitute 17 in place of g(4) to get f(g(4)) = f(17). Then substitute 17 in place of x inf(x) = 2x - 3.Step 4: Following the order of operations we multiply 2 and 17 (34), then subtract 3.

Comparing Graphs of Quadratic Functions:

A parabola is the shape of the graph of


The standard form for the equation of a parabola is (Please remember that f(x) and y can be substituted for each other in quadratic functions.) When changes are made to an equation such as adding a constant, changes in the position of the graph of the equation will take place. If the value of a < 0, the graph of the function will reflect itself across the x-axis. The value of h determines whether the graph will shift right or left and the value of k determines whether the graph will shift upward or downward. See the chart below.

Example 10:

(1) (2) h = -4, k = 7, and a = -1(3) By looking at the chart, we can determine that the graph of should shift to the left 4 units(h = -4), upward 7 units (k = 7), and should reflect itself across the x-axis (a = -1).(4) Compare the actual graphs.


Functions: Absolute ValueThe absolute value of a number is the positive distance between the number and 0 on a number line. Absolute value is denoted by one straight vertical line on each side of a number or expression. Here are some examples of absolute value notation.

(1) |-6| = 6(2) |2| = 2(3) |0| = 0(4) |x - 3|(5) |2x + 6|

The absolute value of the first three examples can be determined because they are all integers. Remember the absolute value of a number is the positive distance between the number and zero on a number line.

An absolute value function is a function which contains at least one absolute value expression. An absolute value function can be evaluated if the value of the variable is known by substituting the value in for the variable and calculating the result.

Evaluating Absolute Value Functions:

Example 1: Evaluate y = |x - 7| + 3, for x = 9.

(1) y = |9 - 7| + 3(2) y = |2| + 3(3) y = 2 + 3(4) y = 5

Step 1: Substitute 9 in place of x in the absolute value function.Step 2: Before the absolute value can be figured, 7 must be subtracted from 9. This is a rule: All operations inside an absolute value symbol must be completed before the absolute value can be taken.Step 3: The distance between 0 and 2 is 2 units, so the absolute value of 2 is 2.Step 4: The final step is to add 2 and 3 to get 5.

Example 2: Evaluate y = 3|x - 2| + |2x + 4|, for x = -3.

(1) y = 3|(-3) - 2| + |2(-3) + 4|(2) y = 3|-5| + |-2|(3) y = 3(5) + 2(4) y = 15 + 2(5) y = 17

Step 1: Substitute -3 in place of x in the absolute value function.Step 2: Compute the values inside both absolute value symbols first.

-3 - 2 = -5 and 2(-3) + 4 = -6 + 4 = -2Step 3: The distance between 0 and -5 is 5 units, so the absolute value of -5 is 5. The distance between -2 and 0 is 2 units, so the absolute value of -2 is 2.Step 4: Three times five is 15.Step 5: Add 15 and 2 to get 17.


Writing an Absolute Value Function as a Compound Function:

Every absolute value function can be written as a compound function. A compound function is a function that is made up of two or more functions. It is very similar to a compound word in language arts. In order to write an absolute value function as a compound function, it is necessary to remember that An example of this rule would be:

When x = 3, |3| = 3; since When x = -3, |-3| = -(-3) = 3; since -3 < 0

Example 3: Write the absolute value function, f(x) = |2x + 6|, as a compound function.

Step 1: In order to write an absolute value function as a compound function, it is necessary to determine the value of x that makes the absolute value function equal to 0. To do this, we set the expression inside the absolute value symbol equal to zero.Step 2: Solve 2x + 6 = 0. Subtract 6 from each side of the equal sign to isolate the 2x on one side of the equal sign.Step 3: Divide by 2 on each side of the equal sign to isolate the x on one side of the equal sign. If -3 is substituted into f(x) = |2x + 6|, f(x) = 0.Step 4: x would be 2x + 6, so -x would be -(2x + 6) = -2x - 6. We use the conditions because -3 is the value that makes f(x) = |2x + 6| equal zero.

Finding the Vertex of an Absolute Value Function:

A vertex is a point at which two line segments, lines, or rays meet to form an angle. The graph of every absolute value function has a vertex. The function f(x) = |x| is graphed below and the vertex is marked.Remember, the terms f(x) and y can be substituted for each other, so f(x) = |x| can be written as y = |x|.


The standard form for an absolute value function is y = a|bx + c| + d. It is not necessary to graph an absolute value function to determine the vertex of the function. If the absolute value function is written in standard form, we can use the standard form to determine the vertex. The vertex of an absolute value function is always written as a coordinate point. The x-coordinate of the vertex is where bx + c = 0. The y-coordinate of the vertex is the value of d.

Example 4: Find the vertex of y = |x - 3| + 2.

(1) x - 3 = 0(2) x = 3(3) d = 2(4) vertex is (3, 2)

Step 1: Set the expression inside the absolute value symbols (bx + c) equal to zero and solve to determine the x-coordinate of the vertex.Step 2: Add 3 to each side of the equal sign to isolate the x and determine that x = 3. The x-coordinate of the vertex is 3.Step 3: Since the 2 in y = |x - 3| + 2 is in the same place as the d in the standard form for an absolute value function, d = 2. The y-coordinate of the vertex is 2.Step 4: Put the two coordinates together to make an ordered pair (x, y). The vertex of y = |x - 3| + 2 is(3, 2).

Example 5: Find the vertex of y = -2|4x + 8| - 6.

(1) 4x + 8 = 0(2) 4x = -8(3) x = -2(4) d = -6(5) vertex is (-2, -6)

Step 1: Set the expression inside the absolute value symbols (bx + c) equal to zero and solve to determine the x-coordinate of the vertex.Step 2: Subtract 8 from each side of the equal sign to isolate the 4x on one side of the equal sign.Step 3: Divide each side of the equation by 4 to get the x by itself on one side of the equal sign. Since


x = -2, the x-coordinate of the vertex is -2.Step 4: Since the -6 in y = -2|4x + 8| - 6 is in the same place as the d in the standard form for an absolute value function, d = -6. The y-coordinate of the vertex is -6.Step 5: Put the two coordinates together to make an ordered pair (x, y). The vertex of y = -2|4x + 8| - 6 is (-2, -6).

Comparing Graphs of Absolute Value Functions:

The standard form for an absolute value function (y = a|bx + c| + d) is needed to compare graphs of absolute value functions with the graph of y = |x|. The variable a in the standard form determines whether the graph opens up or opens down. The variables b and c determine whether the graph shifts to the right or the left and the variable d determines whether the graph shifts upward or downward. See the chart below.

Example 6: Compare the graph of y = |x| with the graph of y = -|2x + 8| - 3.

(1) In y = -|2x + 8| - 3, a = -1, b = 2, c = 8, and d = -3. Refer back to the standard form of an absolute value function to see the placements of a, b, c, and d.

(2) By looking at the chart above, we can see that the graph should open downward (a = -1, which is less than 0). To determine whether the graph shifts left or right, we need to determine c ÷ b = 8 ÷ 2 = 4. Since 4 is greater than zero, the graph will shift to the left 4 units. The last piece of information we have is that d = -3. The chart above states that if d < 0, the graph will shift |d| units downward, so this graph will shift 3 units downward since |-3| = 3.

(3) Compare the actual graphs of y = |x| and y = -|2x + 8| -3.


Matching an Absolute Value Function (Equation) With its Graph:

It is not necessary to graph an absolute value function to match it with its graph. All that is needed is the standard form for an absolute value function: y = a|bx + c| + d.

Example 7: Match the equation y = -|3x - 9| + 2 with its graph below.

(1) The first step in matching the equation of an absolute value function with its graph is to determine the vertex of the graph. The vertex of this graph would be (3, 2) (refer to Example 4 and Example 5 for an explanation about determining the vertex of an absolute value function).

(2) Now that the vertex is determined, we can eliminate any graph that does not have (3, 2) as its vertex. Graphs A and D are eliminated.

(3) The next step in matching the equation to its graph is to look at the value of a in the equation.


In y = -|3x - 9| + 2, a = -1. We learned earlier that if a < 0, the graph would reflect itself over the x-axis (in other words, it will open downward).

(4) Since a = -1 in the equation we are discussing, we can eliminate any graph that does not open downward. Graph B is not the right graph (D would also be out now if it had not already been disqualified in step 2).

(5) The only graph left is graph C and it is the correct answer. If there were still two (or more) graphs left, it would be necessary to substitute some values in for x to determine the value of y and see which of the graphs contained the calculated points.

Irrational Numbers: PiAn irrational number is a number that cannot be written as a fraction. The decimal equivalents of irrational numbers do not terminate (end) and never repeat. For example, 0.10110111011110...and ~ 3.14159265... are two decimals that never repeat and never end.

Determining the Circumference of a Circle:

Circumference is the measurement of the distance around a circle. The circumference of a circle is found by using the following formula:

In this formula, r represents the radius of the circle. The radius of a circle is the length of the segment from the center of a circle to any point on the circle. The radius is one-half the length of the diameter (the distance across a circle) of a circle. Pi is the ratio of the circumference of any circle to its diameter and is approximately equal to 3.14159. Pi is represented by the symbol

Example 1: Find the circumference of a circle with a diameter of 18 inches.

Step 1: Since the radius of a circle is one-half the diameter, divide 18 by 2 to get the value of the radius(9).Step 2: Substitute 9 in place of r in the formula for the circumference of a circle.Step 3: Multiply 9 by 2 (all three terms are multiplied and it does not matter which order numbers are multiplied).

The circumference is 18pi inches.


Example 2: Find the diameter of a circle whose circumference is 87 feet.

Step 1: Substitute 87 in place of C because 87 is the circumference.Step 2: To isolate the r on one side of the equation, divide each side of the equation by Step 3: Since the radius is half the length of the diameter, multiply the radius by 2 to get the length of the diameter. Remember, what is done to one side of an equation must be done to the other side in order for the equation to balance, so multiply the left side of the equation by 2 also. The 2 in the numerator cancels with the 2 in the denominator on the left side of the equation.Step 4: The diameter of the circle equals 87/pi feet.

Finding the Area of a Circle:

The area of a figure is the measure, in square units, of the interior region of a two-dimensional figure. The area of a circle is found by using the following formula:

In this formula, r represents the radius of the circle.

Example 3: Find the area of the circle whose radius is 10.2 inches.

Step 1: Substitute 10.2 in place of r because the radius of the circle is 10.2.Step 2: Square 10.2 (10.2 x 10.2) and multiply the answer by

The area of the circle whose radius is 10.2 is 104.04pi square inches.Example 4: Find the area of the circle whose circumference is 35 inches.

Step 1: Before the area of the circle can be determined, the radius of the circle must be calculated. Since the circumference is known, the formula for the circumference of a circle can be used to determine the


radius of the circle. Substitute 35 in place of C in the circumference formula.Step 2: To isolate the r on one side of the equal sign, divide each side of the equation by Step 3:

Divide 35 by 2 and the radius is Step 4: Now that the radius of the circle is known, it can be substituted in place of r in the formula for the area of a circle

Step 5: Step 6: Square 17.5 (17.5 x 17.5) to get 306.25.

Step 7: The pi in the numerator and one pi in the denominator divide out and leave one pi in the denominator.

The area of the circle, whose circumference is 35 inches, is 306.25/pi square inches.

Finding the Area of an Ellipse:

An ellipse is a closed two-dimensional plane figure that is oval in shape. Every ellipse has two axes. The two axes lie on the symmetry lines and intersect at the center O of the ellipse. One of the axes is the major axis. The major axis contains the foci, has two vertices of the ellipse as its endpoints, and is always longer.

The other axis of an ellipse is called the minor axis. The minor axis does not contain the foci and has two co-vertices of the ellipse as its endpoints.

The formula for an ellipse with axes of lengths 2a and 2b follows.

The length of the major axis is 2a and the length of the minor axis is 2b.

Example 5: Find the area of an ellipse whose minor axis is 23 feet and major axis is 35 feet.

Step 1: The lengths of the axes can be used to determine a and b, so the formula for the area of an ellipse SD Algbra-Questions, Answer Key, Study Guide of 104

can be used. Since the major axis is 35 feet long, 2a = 35. The minor axis is 23 feet long, so 2b = 23.Step 2: Divide each side of each equation by 2 to isolate the variable in each equation.Step 3: Divide 35 by 2 to determine that a = 17.5. Divide 23 by 2 to determine that b = 11.5.Step 4: Substitute the values of a and b into the formula for the area of an ellipse.Step 5: Multiply 17.5 by 11.5 (numbers can be multiplied in any order) and multiply that product by The area of the ellipse is A = 201.25pi square feet.

Example 6: Find the minor axes of an ellipse whose major axis is 25 inches and whose area is 55 square inches.

Step 1: Since the area of the ellipse and the length of the major axis are known, the formula for the area of an ellipse can be used to determine the length of the minor axis. The length of the major axis (25) can be used to find the value of a, so 2a = 25. Divide each side of the equation by 2 to isolate the variable.25 ÷ 2 = 12.5, so a = 12.5.Step 2: Substitute the value of the area (55) and the value of a (12.5) into the formula for the area of an ellipse. We can use this new equation to solve for b (which is 1/2 the length of the minor axis).Step 3: To isolate b on one side of the equal sign, divide each side of the equation by Step 4:

Divide 55 by 12.5 to determine that b = Step 5: Since the length of the minor axis is 2b and we know the value of b, we need to multiply each side of the equation by 2 to determine the length of the minor axis.

Step 6: Multiply b by 2 to get 2b. The length of the minor axis of the ellipse is 8.8/pi inches.

Finding the Volume of a Cone:

A cone is a 3-dimensional figure with one curved surface, one flat surface (usually circular), one curved edge, and one vertex. The volume is the number of cubic units it takes to fill a figure. The formula for the volume of a cone is:

In this formula, r represents the radius of the circle (base) and h represents the height of the cone. (See the diagram of a cone below.)

Example 7: Find the volume of a cone that has a radius of the base equal to 8 inches and height equal to SD Algbra-Questions, Answer Key, Study Guide of 104

12 inches.

Step 1: Since the radius r and the height h are known, substitute the values into the formula for the volume of a cone.Step 2: Following the order of operations, square the 8 first (8 x 8 = 64).Step 3: Multiply 64 and 12, then multiply that product by 1/3. (64 x 12 = 768; 768 x 1/3 = 256)

The volume of the cone is V = 256pi cubic inches.

Example 8: Find the height of a cone that has a radius of the base equal to 10 inches and volume equal to 820 cubic inches.

Step 1: Since the volume of the cone and the radius of the base are known, the formula for the volume of a cone can be used to determine the height of the cone. Substitute the value of the volume and the value of the radius into the formula for the volume of a cone.Step 2: Following the order of operations, square the 10 (10 x 10 = 100).Step 3: To begin to isolate the h on one side of the equal sign, divide each side of the equation by Divide 820 by 100 to get 8.2.Step 4: Multiply each side of the equation by the inverse of 1/3, which is 3/1, to finish isolating the h on one side of the equation. The 3 in the numerator divides out with the 3 in the denominator.Step 5: Multiply 8.2 by 3 get 24.6.

The height of the cone is 24.6/pi inches.

Finding the Volume of a Sphere:

A sphere is a 3-dimensional figure made up of all points in space that are equally distant from a given point called the center. The best example of a sphere is a ball.

The formula for the volume of a sphere is:


In this formula, r represents the radius of the sphere.

Example 9: Find the volume of a sphere whose radius is 11 inches.

Step 1: Substitute the value of the radius (11) into the formula for the volume of a sphere.Step 2: Following the order of operations, calculate Step 3: Multiply 1331 by 4/3. To do this, first multiply 1331 by 4, then divide the product by 3 (1331 x 4 = 5324; 5324 ÷ 3 ~ 1774.67).

The volume of the sphere is V = 1.774.67pi cubic inches.

Example 10: Find the radius of the sphere whose volume is 1,432 cubic feet.

Step 1: Since the volume of the sphere is known (1432), the formula for the volume of a sphere can be used to find the radius of the sphere. Substitute the value of the volume of the sphere into the formula for the volume of a sphere.Step 2: Multiply both sides of the equation by the reciprocal of 4/3, which is 3/4, to begin isolating the r. To multiply 1432 by 3/4, multiply 1432 by 3 then divide that product by 4 (1432 x 3 = 4296; 4296 ÷ 4 = 1074).Step 3: Divide both sides of the equation by pi to isolate the Step 4: Take the cube root of each side of the equation. This will isolate the r on one side of the equal sign.Step 5: Once the r has been isolated, the equation can be written beginning with the r. It is acceptable to leave the radical sign in the answer.

Non-Linear EquationsA non-linear equation is an equation whose graph is not a straight line.


An example of a linear equation (graph is a straight line) is y = x + 3 and an example of a non-linear equation is These two equations are graphed below.

A basic rule to follow to determine whether an equation is linear or non-linear is that non-linear equations have variables that have powers other than one and linear equations have powers equal only to one.

Simplifying Square Roots:

Before we can begin solving non-linear equations, we must discuss how to simplify square roots. When a number is multiplied by itself the product is the square of the number. A square root of a number is a factor that when multiplied by itself equals the number. For example: 2 x 2 = 4, so 4 is the square of 2 also, since 4 = 2 x 2, 2 is a square root of 4. Another square root of 4 is -2 because -2 x -2 = 4. Numbers that have a rational number as their square root are called perfect squares. Examples of perfect squares are: 9 (square root is 3), 16 (square root is 4), 25 (square root is 5) and 9/25 (square root is 3/5).

The notation for a square root is this symbol:

To simplify a square root, we first determine two factors that multiply to make the whole number. One of these two factors should be a perfect square, preferably the largest perfect square that is a factor of the number. Then we take the square root of the perfect square factor and place that number in front of the radical symbol.


Step 1: Rewrite the problem as the square root of two factors of 72. Remember, one of the factors should be the largest perfect square that is a factor of 72. In this case 36 and 2 were used because 36 times 2 equals 72 and 36 is a perfect square.Step 2: Now the problem can be rewritten as the square root of 36 times the square root of 2.Step 3: Determine the square root of 36 (which is 6) and multiply it by radical 2. Since 2 does not have a


factor that is a perfect square, radical 2 does not change.

Example 2: Another way to simplify

Step 1: It is possible to simplify a square root if the largest perfect square factor is not known. Once again, rewrite the problem as the square root of two factors of 72. Make sure one of these two factors is a perfect square. In this case 9 and 8 were used because 9 is a perfect square and 9 times 8 equals 72.Step 2: Since 8 has a factor that is a perfect square (4), the problem must be rewritten as the product of these three factors of 72. 9 x 4 x 2 = 72.Step 3: Since 9 and 4 are perfect squares and 2 does not have a perfect square factor, the problem can be rewritten as the square root of 9 times the square root of 4 times the square root of 2.Step 4: Determine the square root of 9 (which is 3), multiply it by the square root of 4 (which is 2), and multiply them both by radical 2.Step 5: Finally, multiply 3 and 2 to get 6. The 6 is multiplied by radical 2 to obtain the final answer:

Solving Non-Linear Equations:

Non-Linear equations can be solved in much the same way as linear equations. The goal of solving a non-linear equation is to isolate the variable on one side of the equal sign.

Example 3: Solve the following equation.


Answer: Example 4: Solve the following equation.

Using the Pythagorean Theorem to Solve Non-Linear Equations:

The Pythagorean Theorem can be used to determine the length of a missing side of a right triangle. A right triangle is a triangle that has one right (90º ) angle. A 90º angle is marked in a triangle with a box in the angle. The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. The hypotenuse of a right triangle is the side of the triangle opposite the right angle. The other two sides of the triangle are called the legs.


The Pythagorean Theorem states:

In the Pythagorean Theorem, 'c' represents the length of the hypotenuse and 'a' and 'b' represent the lengths of the legs of the right triangle. The Pythagorean Theorem only works for right triangles.

Example 5: Use the Pythagorean Theorem to solve for x.

Step 1: Write the Pythagorean Theorem. Then determine the values of a, b, and c. Remember, c always represents the length of the hypotenuse. In this triangle, the length of the hypotenuse is 8, so c = 8. It does not matter whether a or b is assigned the value of x or the value of 6.Step 2: Substitute the values of a, b, and c into the Pythagorean Theorem.Step 3: Following the order of operations, square the 8 (8 x 8 = 64) and the 6 (6 x 6 = 36).Step 4: Subtract 36 from each side of the equation.Step 5: Take the square root of each side of the equation.Step 6: Simplify radical 28. In this case, the variable represents the length of the side of a triangle, so the answer cannot be negative.

Answer: Solving an Equation by Completing the Square:

Completing the square is a method of solving a quadratic equation in order to express the equation as a single squared term. The method of completing the square is used when an equation cannot be factored. Completing the square involves adding the square of one term to the equation and solving the equation


for the value of the variable. The theorem for completing the square states:

The following is a detailed example of how to complete the square.

Example 6: Solve the equation by completing the square.

Step 1: Add 7 to each side of the equation to put the equation in the form Step 2: Since 8 is in the same place as b, b = 8. Add the square of 1/2 of b to each side of the equation. One-half of 8 equals 4, so we are actually adding 4 squared to each side of the equation.Step 3: We need to fill in the final form in the theorem for completing the square. One-half of b equals 4, so inside the parentheses we have (x + 4). Then we place an exponent of 2 outside the parentheses. Finally, 4 squared equals 16 (4 x 4).Step 4: First, add 7 and 16 to get 23. Then take the square root of each side of the equation. Taking the square root of a term is the opposite of squaring a term, so we get x + 4 on one side of the equal sign. Remember, every number has a positive and a negative square root, so we write Step 5: Subtract 4 from each side of the equation to isolate the x on one side of the equal sign. The new expression cannot be simplified.Step 6: The solution to the equation is

Characteristics that Describe the Graph of a Non-Linear Equation:

A quadratic equation is any equation in the form The graph of a quadratic equation is always a parabola. The vertex of a parabola can be found by putting the quadratic equation in vertex form.


Once a quadratic equation is in vertex form, the vertex is the coordinate point (h, k). If a > 0, then the graph opens up and has a minimum. If a < 0, then the graph opens down and has a maximum.

The roots of a quadratic equation are the solutions of the quadratic equation when y = 0. The roots are the points where the graph intersects the x-axis; therefore, y = 0. The axis of symmetry is the line that passes through the vertex and splits the graph directly in half such that each side is the mirror image of the other. This line is represented by the equation x = h.

Example 7: What are the characteristics of the graph of Step 1: Rewrite the equation in vertex form. The simplest way to do this is by completing the square.

Step 1A: Subtract 4 from each side of the equation.Step 1B: Factor -1 out of the right side of the equation to make the term positive.Step 1C: Add 1/2 of the b term (-6) squared to the right side of the equation and subtract 1/2 of the b term from the left side of the equation (we subtract on the left side of the equation because we factored -1 out of the equation in Step 1B and we need to multiply any number by -1 before we can add or subtract it from the left side of the equation). b = -6, so we are actually adding -3 squared to the right side of the equation and subtracting -3 squared from the left side of the equation.Step 1D: Complete the theorem for completing the square.

Step 2: Determine whether the graph opens up or down. Now that the equation is in vertex form, we can determine that a = -1. Since a < 0, the graph opens down.Step 3: Determine the vertex of the parabola. The vertex is the point (h, k). h = 3 and k = 13, so the vertex is (3, 13).Step 4: Determine the axis of symmetry. The axis of symmetry is the line x = h. The axis of symmetry is x = 3.Step 5: Determine the roots of the equation. The roots of the equation are the values of x when y = 0.


Step 5A: Substitute 0 in place of y.Step 5B: Divide each side of the equation by -1.Step 5C: Take the square root of each side of the equation.

Step 5D: The roots of the equation are If the 13 had been negative, the equation would have had no real roots.

(1) opens down(2) vertex is (3, 13)(3) axis of symmetry is: x = 3

(4) roots are

Rational Numbers: EquationsRational numbers are numbers that can be written as fractions. The following numbers are rational numbers because they can all be written as fractions.

Adding and subtracting rational numbers includes the calculation of whole numbers, fractions, decimals, and integers. It is common for students learning to solve equations with rational numbers to confuse operational symbols with integer signs.

Example 1: Before solving, n - - 7 = - 9, the equation should be simplified to n + 7 = - 9. Solve for n in n + 7 = - 9.

Solution: To solve for n, you need to first isolate n on one side of the equal (=) sign by subtracting 7 from both sides.

Answer: n = - 16

Example 2: Solve for x.


Step 1: Multiply both sides of the equation by the reciprocal of -2/3, which is -3/2. This will isolate the x on one side of the equal sign.Step 2: Make the -6 into a fraction by giving it a denominator of 1 and multiply the two fractions. Remember, multiply numerator by numerator (6 x -3 = -18) and denominator by denominator (1 x 2 =2).Step 3: Reduce -18/2 to lowest terms by dividing -18 and 2 by 2.

Answer: -9

Equations: Addition/SubtractionEquations are number sentences which contain equal signs.

Example: n + 5 = 9

Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified.Example: y - 6

Variables are letters or symbols that represent numbers that are unknown.

The following are examples of how to solve equations.

Example 1: Solve: n + 6 = 20

Step 1: Write the equation.Step 2: Subtract 6 from both sides of the equation.

Answer: n = 14

Example 2: Solve for x.16.6 = 5.5 + x - 8.2

Step 1: Begin to isolate the variable on one side of the equation by adding 8.2 to both sides.Step 2: Subtract 5.5 from both sides to completely isolate the variable.


Answer: x = 19.3

Equations: Multiplication/DivisionEquations are number sentences which contain equal signs:

Example: n + 5 = 9

Expressions are number sentences which do not have equal signs, but need to be evaluated or simplified:Example: y - 6

A common mistake among students first learning to multiply and divide equations is the failure to use inverse operations. For instance, in the equation 3x = 8, the student might multiply 3 to both sides of the equal sign. The correct procedure is to use the inverse operation and divide 3 from both sides (x = 8/3). By applying the inverse operation, the student can isolate the x.

Example 1: Solve: 5n = 75

Step 1: In the equation, solve for the value of n.Step 2: 5n is the same as 5 x n. Divide both sides of the equation by 5.Step 3: 75 ÷ 5 = 15

Answer: n = 15

Example 2: Solve for t.

Step 1: Multiply both sides of the equation by 16 to eliminate the fraction.Step 2: Divide both sides if the equation by 4 to isolate the variable t.

Answer: t = 128.

Inequalities - BAn inequality is a number sentence that uses "is greater than", "is less than", or "is not equal to" symbols. For example, 6n > 4 is a number sentence with an inequality symbol.



Example 1: Solve for y.

8y > 40

Get the variable being solved for (y) on one side of the inequality and the whole number on the other. To do this, divide both sides by 8.

The correct answer is that y is greater than 5.


Example 2: Which inequality represents the value shown on the number line below?

A. n < 3B. n > 3C. n = 3

The answer is A. n < 3 because the dot on the number line is open.

Example 3: Solve the following inequality.-3 < -2x - 3 < 23

Step 1: We can solve by separating the inequality into two inequalities as shown below.-3 < -2x - 3 and -2x - 3 < 23

Step 2: Solve both inequalities.


The answer is -13 < x < 0.

NOTE: -13 < x < 0 is the same as 0 > x > -13.

Equations: Two-StepTwo-step equations require students to perform two operations before solving the equation.

For example, the equation 3x - 3 = 18 requires adding 3 to both sides of the equation and dividing both sides of the equation by 3.

Example 1: Find the value of n in the equation 6n - 9 = 9

Step 1: In the equation, solve for the value of n by first getting n alone. Add +9 to both sides of the equation.Step 2: Divide both sides of the equation by 6.

Answer: n = 3

Example 2: Solve for t.(1/8)t + 3 = 12

Step 1: To get the variable alone, subtract 3 from both sides of the equation.Step 2: Isolate the 't' by multiplying by the reciprocal of 1/8, which is 8/1.

Answer: t = 72


Solving Equations: SubstitutionLetters expressing unknown values in equations are called variables. In these two variable equations, students are given the value of one variable. To find the value of the second variable, substitution must be used.

If y = 4x, and x = 9, substitute the given value of x, in the equation, y = 4x. The result is y = 4(9). Calculate the right side of the equation to get y = 36. Now the values of both variables are known: x = 9 and y = 36.

To check the answer, set up the equation y = 4x as if x was an unknown variable: 36 = 4x. Divide each side of the equation by 4. The result is x = 9, so the values for x and y are both correct.

Example 1: Solve for x.x = 3y + 9y = -2

Solution: Substitute -2 in place of 'y' in the equation. Then, solve for x.x = 3(-2) + 9x = -6 + 9x = 3

When we are not given enough information, the problem cannot be solved.

Example 2: Solve for t.t = s -12r = s + 3

Solution: Since we do not know the value of either 's' or 'r', we cannot determine the value of t.

Exponential Notation - DAn exponent is a number that represents the number of times the base is used as a factor. For example, the number 2 with an exponent of 3 is equal to 2 x 2 x 2. The result of 2 to the 3rd power is 8.

Have the student find the equivalent whole number forms of these exponential numbers:

Scientific Notation:


Scientific notation is based upon exponential properties and is used to communicate very large or very small numbers. Scientific notation deals with significant digits. The most significant digit in a number is the first non-zero digit in the number (reading from left to right). To write a number using scientific notation, place the decimal point to the right of the most significant digit and count the digits between the new placement of the decimal point and the old placement of the decimal point. The number of places that the decimal point moved will be represented by a power of 10.

Example 1: Write 123,000,000 using scientific notation.

Step 1: Determine where the decimal point is in the number to be written using scientific notation.Step 2: Place the decimal point to the right of the most significant digit and count the number of places the decimal point was moved.Step 3: Write all of the significant digits (with the decimal in the new postion) and multiply by 10 to a power. The power on the ten is the number of places that the decimal point was moved. The power is positive because the decimal point was moved from the right to the left.

Answer:

If the decimal point had been moved from the left to the right, the power of the ten would have been negative. To take a number out of scientific notation, move the decimal point the same number of places as the exponent in the power of ten. If the exponent is positive, move the decimal point to the right and if the exponent is negative, move the decimal point to the left.

Example 2: Write the expression in standard form.

Step 1: Rewrite the expression showing the decimal point for the number being multiplied by the power of ten.Step 2: Since the exponent is -4, move the decimal point to the left four places.Step 3: Fill in the places between the decimal point and the significant digit with zeroes. The negative sign that was out in front of the number remains in front of the number and it is negative.

Answer: -0.0007

Evaluating Negative Exponents:

There is a rule for evaluating negative exponents.


Example 3: Evaluate

Step 1: Use the rule for negative exponents to rewrite the expression. The exponent is -3, so we move the (-2) to the bottom of the fraction and make the exponent 3.Step 2: Multiply -2 by itself 3 times to evaluate Step 3: (-2)(-2)(-2) = -8, so the fraction is now 1 over -8Step 4: Rewrite the fraction so that the negative sign is even with the fraction bar.

Answer:

Missing Elements - EEquations will be presented with either a number or operational symbol omitted. Students must solve the equation by determining the missing part.

It may be useful to review the following examples with the student. Help him or her determine whether the missing element is an operational symbol, number, or expression. When the nature of the missing element is determined, make a list of possible answers. When you have the list completed, test each element to determine the correct solution.

Example 1: Find the missing symbol.13 ? 14 = -1

Answer: The missing element is an operational symbol. Possible answers, therefore, are symbols for addition, subtraction, multiplication, or division. The answer is -1, so we can rule out addition and multiplication immediately. The subtraction symbol fits because 13 - 14 = -1.

Example 2: What is the value of N given that A = 87?

10 x N = A - 7

(1) 10 x N = 87 - 7(2) 10 x N = 80(3) 10 ÷ 10 x N = 80 ÷ 10(4) N = 8

Step 1: Substitute the value of A into the equation.Step 2: Calculate 87 - 7.Step 3: Divide each side of the equation by 10 to isolate the missing element on one side of the equation.Step 4: The value of N is 8.


RatesRates refer to the price per unit of a particular commodity. Pricing tactics often cause confusion for today's consumer. Products are sold at 3 for one amount or 5 for another. Calculating rates allows students to figure out the exact cost of the items.

Example: Apples are 5 for $1.25 and bananas are 6 for $1.80, which fruit is less expensive if you only buy one?

(1) $1.25 ÷ 5 = $0.25 --> cost of one apple(2) $1.80 ÷ 6 = $0.30 --> cost of one banana(3) $0.25 < $0.30

Step 1: Determine the cost of one apple by dividing 1.25 by 5. One apple costs $0.25.Step 2: Determine the cost of one banana by dividing 1.80 by 6. One banana costs $0.30.Step 3: Compare the cost of the two pieces of fruit. $0.25 is less than $0.30.

Answer: One apple is less expensive than one banana.

Properties - EMathematical properties are, in essence, statements about the ways in which certain combinations of numbers relate to each other. The student should be familiar with the following properties of multiplication and addition: multiplication by 0, associative, reciprocal, commutative, and order of operations.

Please review the following rules with the student:

1. Multiplication by 0: the product of any integer and 0 equals 0.

-3 x 0 = 0 3 x 0 = 0

2. Associative Property of Addition: (a + b) + c = a + (b + c).

(1 + 2) + 3 = 1 + (2 + 3)

3. Associative Property of Multiplication: (a x b) x c = a x (b x c).

(1 x 2) x 3 = 1 x (2 x 3)

4. Reciprocals: two numbers are reciprocals if their product equals 1.

5. Commutative Property of Addition: a + b = b + a

1 + 2 = 2 + 1

6. Commutative Property of Multiplication: a x b = b x a


1 x 2 = 2 x 1

7. Order of Operations:A. When calculations for a given expression or equation require both addition and multiplication, the rule is to multiply first and add second.

(3)(2) + 3 = ?6 + 3 = ?6 + 3 = 9

B. The number outside the parentheses is multiplied with each number within the parentheses:x(y + z) = xy + xz.

(1) 3(x + y)(2) 3(x) + 3(y)(3) 3x + 3y

C. If a given expression contains both parentheses and brackets, calculations should be completed working from the innermost parentheses or bracket outward.

The following are sample questions using the above properties.

Example 1: Which answer best completes the number sentence?5 = (5 x 4) + (5 x 6)

A. x (4 + 6)B. + (4 + 6)C. x (20 + 30)D. + (4 x 6)

Answer: A (because of rule 7A)

Example 2: Which answer best completes this number sentence?0.32 + 0.45 + -62 = ?

A. -0.15B. 0.32C. 0.15D. 0.45

Answer: C


Example 3: Which one of the following best completes the number sentence?(2.3 + 3.1) x 5.6 = ?

A. 3.5 x 5.6B. 7.2 x 5.6C. 9.1 x 5.6D. 5.4 x 5.6

Answer: D (because of rule 7B)

Example 4: What is the value of n in the following statement?13 x (3.4 x 0) = n

A. 44.2B. 0C. 13D. 3.4

Answer: B (because of rule 1)

Radicals: SimplifyingA radical sign looks like a check mark with a line across the top. The radical sign is used to communicate square roots. The numbers -3 and 3 are the square roots of 9 because both -3 and 3 squared equal 9. The definition of a square root is: if x squared = y, then x is a square root of y. Simplifying radicals assesses the ability to perform operations with radicals.

In order to simplify radicals, it is important to understand the product property of square roots. We already know the square root of 81 is 9. We can also determine this by the following method:

Simplify the square roots below:

When simplifying with variables, we can do the following:

Also, the same method can be used to simplify radicals taking the 3rd, 4th, 5th,...etc. roots.


Functions/Relations - BA relation can be expressed as a set of ordered pairs such as A function is a special type of relation.

There are a few ways to determine if a relation is a function. One way is to look at all of the ordered pairs of a relation. If no two of these ordered pairs have the same x - term (abscissa), then the relation is a function. Another method involves graphing the ordered pairs of a relation on a coordinate graph. If no vertical line crosses this graph at more than one point, then the relation is a function.

A linear function is a function whose graph is a line or subset of a line which is not vertical. A special type of linear function is a constant function, whose graph is a horizontal line or subset of a horizontal line.

Using these guidelines, find the value for q that will make the following relation not a function:

The answer is 4 because if g = 4, then the second ordered pair would be (16, 7). The two ordered pairs would have the same x - term, and for that reason the relation would not be a function.

Equations: SystemsA system of equations contains at least two equations that may be linear, non-linear, or a combination of the two types. A graphical interpretation of the solution of a system of equations is that point (or points) where the graphs of the equations intersect. One method of finding the solution(s) of a system of equations involves adding the two equations together.

Example 1:


Step 1: Write the equations in a vertical format, aligning the x-terms, y-terms, equal signs, and constant terms.Step 2: The objective is to add the corresponding parts of the two equations together and eliminate either the x- or y-term. Multiplying each term in the top equation by 4 will create the necessary conditions for eliminating the y-term.Step 3: Add like terms in the two equations, and notice the resulting y-term will have a coefficient of zero and be eliminated.Step 4: Solve the resulting equation for x. In this case, that means divide both sides of the equation by 14. This results with x = 5.Step 5: To solve for y, substitute the value of x (5) in either of the original equations. The second equation was chosen for this example.Step 6: Subtract 10 from both sides of the equation.Step 7: Solve the resulting equation for y. In this case, that means divide both sides of the equation by -4. This results with y = -2.

As with all other equations, substitute the values of x and y into the original equations to ensure they are correct solutions.

Both values check out. Therefore, the solution to the system of equations are x = 5 and y = -2. This can be interpreted as the ordered pair (5, -2) of the point of intersection of the graphs of these two equations.

Equations of a LineEvery line on any coordinate graph has a corresponding equation which describes every point on the line. Every linear equation (equation of the line) contains a slope. The slope of a line is the same between any two points on the line.

Before you can find the equation of a line, you must first be able to find the slope of a line when given two coordinate points on the line. These two points are named: The formula for the slope of a line follows.


Example 1: Find the slope of the line between Point R (2, 4) and Point S (1, 3).

Step 1: Substitute the given coordinate points into the formula.Step 2: Simplify the fraction.

Answer: The slope of the line is 1.

The Point-Slope form for the equation of a line:

Example 2: Use the following points to find the equation of the line.Point T (7, -3)Point U (-4, 6)

Step 1: Solve for the slope of the line between Point T and Point U.Step 2: Use one of the coordinate points and the slope and substitute them into the Point-Slope form for the equation of a line.Step 3: Simplify both sides of the equation.Step 4: Subtract 3 from both sides of the equation.

The equation of the line that passes through (7, -3) and (-4, 6) is y = -9/11x + 30/11.

Number Relation ProblemsNumber relation problems involve sentences that must be translated into equations.

To translate sentences, review the following phrase/number equivalences:

If x is equal to 4, then x = 4If the sum of 2x and 2y is 23, then 2x + 2y = 23If 3 is less than twice y, then 2y > 3If 5x decreased by y is more than 2y, then 5x - y > 2y If the result of 7 more than 3 times x is y, then 3x + 7 = y


Example: One number is 2 less than another number. If twice the larger number is decreased by 3 times the smaller number, the result is 20.

Step 1: Translate the first sentence in the problem into the first equation.Let x and y represent the unknown numbers, with y representing the larger number and x representing the smaller number.Step 2: Translate the second sentence in the problem into the second equation.Use x and y as in the first equation, substituting them into their respective locations in the equation.Step 3: Substitute x + 2 for y in the second equation.Step 4: Distribute the 2 across (x + 2).Step 5: Combine the similar terms.Step 6: Subtract 4 from both sides.Step 7: Divide each side by -1 to find the value of x.Step 8: Using x = -16, substitute for x in the second equation.Step 9: Multiply -3 and -16.Step 10: Subtract 48 from both sides.Step 11: Divide each side by 2.

Answer: x = -16 and y = -14

As always, check the solutions by substituting both equations with x = -16 and y = -14.

FactoringConsider the following equation:

3 x 4 = 12The numbers 3 and 4 are said to be factors of the number 12. This concept of factoring is not reserved for numbers, but may be extended to polynomials as well.

Factoring is the breaking up of quantities into products of their component factors. One way to think of factoring is as the opposite or inverse of multiplying.


A polynomial is a term or sum of terms. Each term is either a number or a product of a number and one or more variables.

A monomial is a polynomial with one term.

A binomial is a polynomial with two terms.

A trinomial is a polynomial with three terms.

Consider the following polynomial:

A typical question on factoring will include a polynomial like the one above. Notice that 4y is a common factor of each term of the polynomial.

Step 1: Factor out the 4y by dividing each term of the trinomial by 4y.

Step 2: The trinomial in parentheses can be factored further. Since the coefficient of the "y squared" term is equal to 1, focus on the last term, in this case, -5. If factors of -5 can be found that ADD up to the coefficient of the middle term (in this case, 4) the trinomial can be factored. Two factors of -5 are 5 and -1, and when ADDED together the result is equal to the coefficient of the middle term, 4. Notice how these numbers are put together to construct the fully factored trinomial:

(4y)(y + 5)(y - 1)

To check the result, use the rules for multiplying polynomials and you should have the original polynomial when finished.

Sometimes factoring must be done by grouping. The polynomial given below may appear impossible to factor at first, but if you examine the steps you will see a method to use with polynomials of this type.

Step 1: (5x) is a common term to both 15x and 20xy. Factor it out of only those two terms.Step 2: (6n) is a common term of both 18nx and 24ny. Factor it out of only those two terms.Step 3: Notice that the quantity (3x + 4y) is a common factor of 5x and 6n. The expression is rewritten to indicate this, and the polynomial is completely factored.

A special type of polynomial expresses the difference of two perfect squares. Polynomials of this type are factored easily once the rule is remembered.


Since each term in the polynomial is a perfect square, the square root of each term (in this case x and 6 respectively) will be used in the following way. The original polynomial is factored as (x + 6)(x - 6). Notice that if these terms are multiplied together, the original polynomial is formed. Polynomials that are in the "difference of squares" form may always be factored as the sum of the square roots times the difference of the square roots.

Quadratic FormulaA quadratic equation is a polynomial equation in which the highest power of the unknown variable is two.

An example of a quadratic equation is below.

The format of a quadratic equation is Quadratic equations can be solved by factoring, graphing, or by using the quadratic formula. The quadratic formula is as follows:

It can be found in any algebra textbook. This formula should be memorized.

To apply the formula to a quadratic equation, use the quadratic equation format given above as a guideline.

Example 1: Solve the quadratic equation.

Step 1: Determine the values of a, b, and c and substitute them into the quadratic formula. a = 1, b = 6, and c = -91Step 2: Determine the value under the radical symbol. 6 squared is 36 and -91 times -4 equals 364.36 + 364 = 400Step 3: The square root of 400 is 20 (20 x 20 = 400).Step 4: Split the remaining problem into two problems: (-6 + 20) ÷ 2 and (-6 - 20) ÷ 2 and solve the two problems.


The answers are x = 7 and x = -13.

Example 2: Solve the quadratic equation.

Step 1: Write the equation.

Step 2: This will put the equation in standard form.Step 3: Determine the values of a, b, and c.Step 4: Substitute the values of a, b, and c into the quadratic formula.Step 5: Determine the value under the radical sign. The square root of 0 is 0.Step 6: Solve for x.

Answer: x = -2

The discriminant is the portion of the quadratic equation under the radical sign The discriminant properties below will give you vital information about quadratic equations.

1. If the discriminant is a perfect square, then the quadratic equation can be factored.2. If the discriminant is greater than 0, then the equation has two real solutions.3. If the discriminant is less than 0, then the equation has no real solutions.4. If the discriminant is equal to 0, then the equation has one real solution.

Example 3: How many solutions does the following quadratic equation have?


Step 1: Determine the values of a, b, and c.Step 2: Substitute the values for a, b, and c into Step 3: Simplify the discriminant.

Since the discriminant is greater than zero, there are two real solutions.

If the discriminant is a perfect square, then the solutions are rational.If the discriminant is not a perfect square, then the solutions are irrational.

Polynomials: AdditionA monomial is the product of a number and an unknown variable or unknown variables. 6xy is a monomial. The sum of two or more monomials is called a polynomial. Here is an example of a polynomial: Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are monomials that have the same variable or variables for which the variable or variables have the same exponent.

To add polynomials, combine similar terms.

Example 1:

Step 1: Set the two polynomials up as one long polynomial.Step 2: Combine the like terms.Step 3: Add the results of combining the like terms to determine the answer.

The sum is Example 2:

Step 1: The area of the large rectangle would be the area of the shaded region added to the area of the


small rectangle. Since we know both areas, we simply add them together.Step 2: Collect like terms so they can be added.Step 3: Add together any like terms that were collected to determine the final answer.

The area of the large rectangle is Sometimes it is necessary to use the distributive property before we can combine like terms.

Example 3:

Step 1: Multiply each term of the first polynomial by 3. Then multiply each term of the second polynomial by -4.Step 2: Group like terms together.Step 3: Combine like terms.

Answer: Example 4: Solve for a, b, and c.

Step 1: Group the similar terms on the left side of the equation together.Step 2: Now, group like terms from both sides of the equal sign together.Step 3: Solve for a, b, and c.

Answer: a = -4, b = -2, and c = 1

Exponential Notation - EExponents communicate the number of times a base number is used as a factor. The base number 5 to the 3rd power (with an exponent of 3) translates to 5 x 5 x 5. (5 to the 3rd power is not 5 x 3.) The result of 5 to the 3rd power is 125. To perform operations with exponents, all exponential properties must be understood.

To simplify an expression or to find the missing term in a simplified expression, apply the following exponential properties which are listed below.

Product of Powers:SD Algbra-Questions, Answer Key, Study Guide of 104

When multiplying two (or more) numbers with the same base that have exponents, the base remains the same and the exponents are added.

Power to a Power:

When taking a number with an exponent to another power, the base remains the same and the exponents are multiplied.

The Power of Zero:

Any number taken to the power of zero (except zero) equals 1.

Negative Exponents:

There is a rule for evaluating negative exponents.


Step 1: Separate the expression into products of whole numbers times products of variables with like bases. A variable or number that is written without an exponent automatically has an exponent of 1.Step 2: First, multiply the whole numbers to get 6. Use the "Product of Powers" rule to evaluate the terms in the second set of parentheses and the terms in the third set of parentheses.Step 3: Add the exponents to complete the expression.

Answer:



Step 1: Separate the expression into products of each of its terms. The exponent (2) is given to each term inside the parentheses.Step 2: -2 to the second power is (-2)(-2). Use the "Power to a Power" rule to evaluate the rest of the expression.Step 3: Multiply the powers to determine the exponents on the variables. The final expression can be written with the parentheses still in it, or without the parentheses. It is more acceptable to remove the parentheses.

Polynomials: MultiplicationA monomial is the product of a number and an unknown variable or unknown variables. 6xy is a monomial. The sum of two or more monomials is called a polynomial. Here is an example of a polynomial: A binomial is a polynomial with exactly two monomial terms. 3x + 4 is a binomial. A trinomial is a polynomial with exactly three terms. 4xy - 3x +6y is a trinomial.

Adding and subtracting polynomials includes simplifying and combining "like" terms. Like terms are monomials that have the same variable or variables for which the variable or variables have the same exponent.

To multiply monomials and polynomials, the exponential properties must be followed. These properties apply to all real numbers with positive exponents.

Exponential property #1 can be used to obtain results for problems such as:

Exponential properties #2 and #3 can be used to obtain results such as:

To multiply a monomial and a polynomial, multiply the monomial by each term of the polynomial. In other words, distribute the monomial.


To multiply two polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial. Distribute each term of the first polynomial across the second polynomial.

Example 1: Which choice below represents the following trinomial:

We know that when multiplying two polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial. Let's look at our solutions.

Answer: B

Example 2: Given these measurements, find the area of the rectangle.

Step 1: Area = Length x Width. Determine the length and width of the rectangle.Step 2: Multiply the length and width to determine the area of the rectangle.Step 3: Use FOIL to multiply the two binomials. Remember, multiply the first terms in each binomial, then multiply the outer terms, next multiply the inner terms, and finally multiply the last terms.Step 4: The final step in solving this problem is to add the like terms. 3n + 7n = 10n

Answer:

Polynomials: DivisionA monomial is the product of a number and an unknown variable or unknown variables. 6xy is a monomial. The sum of two or more monomials is called a polynomial. Here is an example of a polynomial:


A binomial is a polynomial with exactly two monomial terms. 3x + 4 is a binomial. A trinomial is a polynomial with exactly three terms. 4xy - 3x +6y is a trinomial.

Before dividing polynomials, recall the following properties associated with exponents:

Example 1: Divide.

Step 1: Divide the whole numbers: 12 ÷ -3 = -4.Step 2: Use the properties above to divide the variables. Begin with the x-variables. x-cubed divided by x equals x-squared.Step 3: Now divide the y-variables. y divided by y equals y to the power of zero. Any number taken to the power of zero equals 1.Step 4: Finally, multiply the quotients back together.

The answer is Dividing a Polynomial by a Monomial:

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. Then, combine the similar terms.

Example 2: Divide.

Step 1: Divide 3m by 3, to get m.Step 2: Divide -9n by 3, to get -3n.Step 3: Combine the terms.

Answer: m - 3n

Dividing a Polynomial by a Polynomial:


Dividing one polynomial by another is very similar to long division.

Example 3:

Step 1: Write the problem as a long division problem. The binomial belongs on the outside of the division symbol because it is the term we are dividing by.Step 2: Now, we can begin dividing.

So, 2x belongs above the 8x.Step 3: The next step is to multiply 2x by (3x + 1).

Subtract that product from Now, bring the + 8 straight down beside the 6x.Step 4: (3x)(2) = 6x, so we place the 2 above the 8 in the answer.Step 5: Multiply 2 by (3x + 1) to get 6x + 2. Subtract (6x + 2) from (6x + 8). There is a remainder of 6, so we write the remainder as a fraction with the binomial as the denominator.

Answer:

Sets/Subsets/Solution SetsA set is a collection of numbers. A subset is a collection of numbers that all belong to a larger set. Every set is also a subset of itself. If Set C is a subset of Set D, then every element of Set C is also in Set D. Set C is a part of (a subset of) Set D. A solution set of an open sentence is the set of all elements which make the sentence true when replaced for the unknown.

Sets are indicated by braces, Example 1: Find the solution.

2x + 3 = 3(x + 3)


Step 1: Distribute the 3 to the numbers in the parentheses. This involves multiplying each term in the parentheses by 3.Step 2: We want the variable on one side by itself. Subtract 2x from each side of the equation to get the variable on one side of the equation.Step 3: Subtract 9 from each side of the equation to isolate the variable on one side of the equation.Step 4: Rewrite -6 = 1x with the variable first. x = -6

Answer: x = -6

Example 2: Choose the corresponding number line: y < -2 or y > 3.

The answer is B.

We are also able to determine sets and subsets from Venn diagrams. Venn diagrams are drawings that show relationships among sets of items. Venn diagrams are usually sets of overlapping circles or ovals. The relationships between the sets can be represented by the following symbols.

Example 3: Given the sets A, B, and C expressed in the diagram, which option is NOT true?

Step 1: Determine the numbers that are in each set.

Step 2: Use the sets to determine which of the four answer choices is NOT true.


Answer: Choice D is NOT true.

Absolute Value: SolveThe absolute value of a real number is the distance the real number x is from 0 on a number line. The absolute value of a real number is denoted by placing the real number within two vertical lines: |x|. In other words, |-5| ("the absolute value of -5") is 5 because -5 is 5 units from 0 on a number line.

There are two major principles of absolute value:1. The absolute value of a negative number is always a positive number.2. The absolute value of 0 is 0.

One way to find the absolute value of single digit numbers such as |-5|, is to draw a number line and plot -5 and 0 on the number line.

Notice that -5 is 5 points from 0, so the absolute value of |-5| is 5.

The student must also be able to find the absolute value of single digit numbers, solve absolute value equations, evaluate absolute value expressions, and recognize absolute value expressions on number lines.

The following example demonstrates how to solve equations with absolute value.

Example 1: Solve: |2x - 3| = 17

Step 1: Set the equation up to equal both 17 and -17. This is because 2x - 3 can equal either 17 or -17 to make the original equation true.Step 2: Solve the equation 2x - 3 = 17. Add 3 to both sides of the equation to get 2x = 20. Then divide each side of the equation by 2 to get x = 10.Step 3: Solve the equation 2x - 3 = -17. Add 3 to both sides of the equation to get 2x = -14. Then divide each side of the equation by 2 to get x = -7.

The solutions for |2x - 3| = 17 are x = 10 or x = -7. However, to be sure, always check solutions by substituting both solutions (10 and -7) for x (in |2x - 3| = 17). For x = 10 (shown in Step 4) and forx = -7 (shown in Step 5):


Example 2: Evaluate expression |x - 3| when x = -2.

(1) |-2 - 3|(2) |-5| = 5

Step 1: Substitute -2 in the absolute value expression for x.Step 2: The expression |-2 - 3| becomes |-5|. The absolute value of |-5| is 5 because -5 is 5 points from 0 on a number line.



Example 3: Solve and graph the inequality. |3x + 5| < 10

Step 1: Set up the inequality as being less than 10 and greater than -10. (The second inequality should have the "is greater than" symbol because: when we make the 10 negative, we turn the inequality symbol the opposite direction. In this case, we change it from "is less than" to "is greater than.")Step 2: Solve the inequality 3x + 5 < 10. Subtract 5 from both sides of the inequality, then divide both sides of the inequality by 3.Step 3: Solve the inequality 3x + 5 > -10. Subtract 5 from both sides of the inequality, then divide both


sides of the inequality by 3.

The answer is: x < 5/3 and x > -5, which can also be written as -5 < x < 5/3.

Now, graph the inequality. Place an open dot on the -5 and an open dot on 5/3 because the inequality is strictly less than 5/3 and strictly greater than -5 (x cannot equal 5/3 or -5). Shade the portion of the graph that is greater than -5 and less than 5/3.


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