Unit 5 Systems of Equations or Inequalities

In analyzing data and problem solving in general there will be times when we willput two lines on one set of coordinate axes. Below we see a graph of the monthly expenses and income for Great Western Jeep Tours. At the beginning of each monththey must pay a number of bills or fixed expenses. These bills include vehicle payments, property rental, insurance, utilities and other expenses. During the month they will have income for each tour they give to customers. During the month they must pay their employees for each trip and fuel for each tour. At the start of each month the business will have no income, but they will have to pay the fixed expenses.

Great Western Jeep Tours Income and Expenses

The solid line in the graph represents the income for the month. The income goes upwith each tour the company gives to a customer. The dashed line represents the expenses the company has during the month. The expenses start at over $6000 afterpaying all the fixed expenses. The expenses then go up when the company pays for the fuel and the wages for the guides.

One of the most important points on the graph is where the two lines intersect or cross.The place where the two lines cross is a point that makes both equations true. It is also the point where the two equations have the same y-value. The y-values of the solid lines are the income. The y-values of the dashed line are the expenses. When

0 10 20 30 40 50 60 70

Tours

$21000

$18000

$15000

$12000

$9000

$6000

$3000

$0

Expenses

Income

the y-values are equal the income and the expenses are equal. To the left of this point the graphs show the business losing money to the right they are making money.This is the point at which the business breaks even. The business must have give about 23 tours to break even. Finding this point is important to any business. In thisunit we will be finding a place where functions share a point.

Unit 5 Vocabulary and ConceptsParallel Lines Lines that are in the same plane and do not intersect are

called parallel lines.

Perpendicular Lines Lines that intersect and create right angles are called

perpendicular lines. System A system is two equations or inequalities.

Key Concepts for Systems of Equations or Inequalities

Points on a Line A point is on a line if its coordinates, when substituted into the

equation, makes it true.

Point of Intersection A point of intersection is the place where two lines cross. This point is the solution to the system and its coordinates

will make both equations true.

Method of Substitution A way to solve a system of equations by replacing one

of the variables in an equation by the expression it is

equal to from a second equation.

Method of Elimination A way to solve a system of equations by adding or

subtracting two equations.

Objective The student will solve systems of equations graphically.

A system of equations consists of two or more functions. In Algebra I we will studysystems of two linear equations. Before we examine two equations we must understand how we can tell when a point is one a line and when it is not. If we were to graph the equation y = 2x - 4 as illustrated below we would find the slope and intercept, then graphthe line. In the diagram below point A(3, 2) is on the line and point B(1, 6) is not on the line.

y = 2x – 4

The conclusion or conjecture we can draw from this is that “If a point is on the line then the point will make the equation true.” We can use this conjecture to determinewhen a point is on a line. The examples below illustrate this idea.

Example A Given the equation of a line y = .5x + 3.5 is (4, 2) on the line?We substitute the point into the equation y = .5x + 3.5

2 = .5(4) + 3.5 2 = 2 + 3.2 2 = 5.5

This is false and the point is NOT on the line.

Example B Given the equation of a line y = -4x + 1 is (-1, 5) on the line?We substitute the point into the equation y = -4x + 1

5 = -4(-1) + 1 5 = 4 + 1

Unit 5 Section 1

A

B

If we substitute point A into the equation we get y = 2x – 4

2 = 2(3) - 42 = 6 – 42 = 2

which is true!

If we substitute point B into the equation we get y = 2x – 4

6 = 2(1) - 46 = 2 – 46 = -2

which is false!

5 = 5This is true and the point is on the line.

When we examine a system of equations we will be looking for the point at which the two lines intersect or cross. As we saw in the “Unit 5 Introduction” this is an important point in applications. The point at which two lines intersect is the onlypoint that will make both equations true. We can illustrate this with the example below.

The graph below has the lines y = x + 1 and y = -2x + 4. We will test the pointsA(-1, 0), B(1, 2), and C(4, 1).

The conclusion we can draw from the example above is:

B

AC

When we substitute point A into the equations we get:

y = x + 1 y = -2x + 40 = -1 + 1 0 = -2(-1) +

40 = 0 0 = 2 + 4

0 = 6Point A only makes one of the equationstrue and it is NOT on both lines.

When we substitute point B into the equations we get: y = x + 1 y = -2x + 4 2 = 1 + 1 2 = -2(1) + 4 2 = 2 2 = -2 + 4

2 = 2Point B makes both of the equationstrue and it is on both lines. It is the point where the lines intersect.

When we substitute point C into the equations we get: y = x + 1 y = -2x + 4 1 = 4 + 1 1 = -2(4) + 4 1 = 5 1 = -8 + 4

1 = -4Point C makes both of the equationsfalse and it is Not on either line.

If a point makes both equations true then the point is where the two lines intersect.

There are several methods for finding the point of intersection. We will begin by examining the graphical method. This method is similar to what we have seen in the example above. This consists of graphing the two lines and finding the point of intersection by inspection. The point can be checked by substituting the coordinates into the original equations. Examples of this process are on the following page.

Example A Given the equations y = 3x – 6 and y = -x + 2 find the point of intersection.

In order to graph the two lines we need to find their slopes and y-intercepts.

The first equation y = 3x – 6 has m = 3 and b = -6. The second equation

y = -x + 2 has m = -1 and b = 2. They two lines are graphed below.

The point of intersection is wherethe two points cross. The coordinates of the point are

(2, 0)

We can check our answer by substituting the point into both equations.y = 3x – 6 y = -x + 20 = 3(2) – 6 0 = -(2) + 20 = 6 – 6 0 = -2 + 20 = 0 0 = 0

The point makes both equations true and so is the solution to the system.

Example B Given the equations y = 2x + 1 and y = -.5x - 3 find the point of

intersection. In order to graph the two lines we need to find their slopes

and y-intercepts. The first equation y = 2x + 1 has m = 2 and b = 1. The second

equation y = -.5x - 3 has m = -.5 and b = -3. The two lines are graphed below.

The point of intersection is wherethe two points cross. In this example we must estimatecoordinates of the point

(-1.3, -2.1)

Since this is an estimate checking it will not result in equations that are necessarily true.

VIDEO LINK: Khan Academy Solving Linear Systems by Graphing

Exercises Unit 5 Section 1 Set A 1. The point where two lines intersect will make the two equations ______________.

For each given system of equations check to see if the listed point is the solution.Answer yes or no and show your work.

2. y = 3x - 4 3. y = -x + 4 4. y = .5x + 3 y = 2x - 3 y = 3x + 8 y = 3x - 2

(1, -1) (2, 2) (2, 4)

5. 2y + 4x = 0 6. y = x + 1 7. x = y + 2 y - x = 3 x + y = 11 3y = x - 5

(-1, 2) (6, 5) (3, 1)

Graph each pair of lines and find the point of intersection.

8. y = x - 1 9. y = - x - 4 10. y = 32 x - 1

y = -2x + 8 y = x + 5 y = -2x - 1

11. y = -5x + 8 12. y = x + 1 13. y = 2 y = 3x - 8 y = 2x + 4 x = -3

14. y = -x - 3 15. y = x + 4 16. y = 2.5

y = x + 1 y = - x - 2 y = x

17. Given the data set below find the mean, median, mode, and range.12, 14, 11, 17, 13, 14, 15, 9, 10, 14, 11

18. The data set below represents the ages of people in a bowling league. Create a stem and leaf plot to represent the data.

28, 33, 42, 25, 36, 31, 33, 45, 40, 46, 27, 38, 51, 33, 49

19. Given the graph of the line to the right represents the first equation is a system.

a. Graph the line y = x - 10

b. Find the solution to the system of equations. Explain how you identified the solution.

c. If the equation for the given line is y = - x + 2, prove you found the correct solution using substitution.

20. Given the graph of the line to the right. Which of the graphs below shows the point of intersection of the given line with y = -x - 2.

a) b)

c) d)

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Exercises Unit 5 Section 1 Set B

1. When a single point can make two different equations true we know this is a point at which the two lines __________________.

For each given system of equations check to see if the listed point is the solution.Answer yes or no and show your work.

2. y = 5x - 7 3. y = -2x + 4 4. y = -x + 4 y = 2x + 2 y = x - 10 y = 4x - 2

(3, 8) (5, -6) (-2, 6)

5. 2y + .5x = 0 6. y = x + 3 7. x = 2y y - 2x = 4 x - y = -4 -y = x - 10

(8, -2) (-2, 2) (18, -9)

Graph each pair of lines and find the point of intersection.

8. y = 2x - 4 9. y = x + 1 10. y = - x - 1 y = -3x + 6 y = x + 3 y = x - 4

11. y = -3x + 4 12. y = x 13. y = -4 y = x + 4 y = 2x x = 1

14. y + x = 0 15. y = x + 4 16. y = 2x + 1 2y = 4x + 6 y = x - 2 2y = 4x + 2

16. The solution to the equation -2(x – 4) + x = x – 6 is

a. 1 b. 7 c. -1 d. 14

17. The solution to the inequality -3x + 1 > 7 is

a. x > -2 b. x < c. x< -2 d. x >

19. Given the graph of the line to the right represents the first equation is a system.

a. Graph the line y = x - 4

b. Find the solution to the system of equations. Explain how you identified the solution.

c. If the equation for the given line is y = 3x + 6, prove you found the correct solution using substitution.

20. Given the graph of the line to the right. Which of the graphs below shows the point of intersection of the given line with y = .5x + 11.

a) b)

c) d)

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Objective The student will define, recognize and find parallel and

perpendicular lines.

There are two important ways lines can relate in a plane. These are called parallel and perpendicular.

Below is an example of what these lines would look like with and without a graphing plane.

Because the lines go in the same direction they will never intersect. These kinds of lines appear everywhere in our world. The sides of a window, the legs of a table, the edges of a book, a railroad track, and many other figures or objects have parallellines in them. Also because the lines go in the same direction they have equal slopes.This idea will make it possible for use to identify and draw parallel lines.

Below is an example of how perpendicular lines would look.

The angles formed here are 90o and are called right angles. These angles are the

Unit 5 Section 2

DefinitionParallel lines are in the same plane and do not intersect.

DefinitionPerpendicular lines form right angles when they intersect.

Both of these lines have slopes of

“right” angles because when we need to build walls or use rectangles or squares these are the angles we must use. So they are the “right” angles for many purposes. When we see angles like these graphed on the x-y plane we can see that they go in opposite directions. There are two perpendicular lines graphed in the next diagram.

An Example of Perpendicular Lines

Another way to see how perpendicular lines are created is to start with a vertical lineand a horizontal line and rotate them as in the diagrams below.

Perpendicular Lines Rotated 15o Rotated 45o

If we examine the slopes in these diagrams we can draw some conclusions. Top Diagram Rotated 15o Diagram Rotated 45o Diagram

m = 2 and m = - m = 3 and m = - m = 1 and m = -1 or -

When we look at the pairs of slopes that produce perpendicular lines we find they are opposite in sign and they are reciprocals. This conclusion makes it possible for us to identify pairs of perpendicular lines from their equations and to draw perpendicularlines by counting the slopes.

There are many symbols we use in Algebra. We have symbols for the two new relationships we now have. For parallel lines we use two vertical bars that look like parallel

These two lines are perpendicular. They go in opposite directions. This means thatone line has a positive slope and the otherhas a negative slope. One of the lines is very steep while the only slants a little. that is why we can say they go in opposite

lines. The symbol is ||. For perpendicular lines we use a symbol that looks like the intersection of two perpendicular lines. The symbol is .

The examples that follow show us how to draw || and lines.

Example A Given the graphed line draw a || (parallel) line through the point (3,2). Original Graph Parallel Line

Example A Given the graphed line, draw a (perpendicular) line through the point (-2,3). Original Graph Perpendicular Line

We also need to be able to recognize when lines are parallel or perpendicular when

we are given their equations. The ideas we can use to identify these relationshipsare given below.

Below are examples of how to use this fact to recognize parallel lines. Example A Example B Example C Given the equations Given the equations Given the equations

y = 3x + 1 2x + y = 7 y = -5x + 1y = 3x – 5 -4x – 2y = 10 y = 5x - 4

By counting we know the original line has a slope of m = -2

So we need to plot the point (3,2) and then count the slope of -2 to find points

By counting we know the original line has a slope of m = The opposite reciprocal is m = -So we need to plot the point (-2,3) and then count the slope to find points for the new line.

Lines are parallel when their slopes are equal.

y = -2x + 7 y = -2x – 5

Below are examples of how to use this fact to recognize parallel lines. Example A Examples B

Given the equations Given the equations y = x – 1 y = x + 4

y = -2x + 4 y = - x - 2

We may also be asked to use the relationships of parallel and perpendicular to solveother kinds of problems. Some examples follow. Example A Find the equation of a line parallel to y = 6x – 4 passing through the

point (-1, 3).

Since the slope of the given line is m = 6 the problem becomes something like what we did in Unit 4. We need to find the equation of a line with a slope of m = 6 and passing through (-1, 3). So we follow our process:

y = m x + b Write the slope-intercept form of an equation. 3 = 6(-1) + b Substitute the coordinates and slope. 3 = -6 + b +6 +6 9 = b

Both equations have a slope of m = 3.So the two lines areparallel.

We must first isolate yin both equations. Weget the equations below.

The slopes are equal,the lines are ||.

The slopes are not equal.The lines are not

Lines are perpendicular when their slopes are opposite reciprocals.

The slopes are opposite in sign. The slope of m = has a reciprocal of m = = 2 Since the slopes are opposite reciprocals the lines are .

The slopes are opposite in sign. The slope of m = has a reciprocal of m = Since the slopes are opposite reciprocals the lines are .

So our equation is y = 6x + 9, which is the equation of the line.

Example B Find the equation of a line to y = x + 1 passing through (2, -4).

The slope of the given line is m = and the opposite reciprocal is m = -3.

Again we will follow the process for finding the equation of a line with a slope of m = -3 and passing through (2, -4).

Example B (continued)y = m x + b Write the slope-intercept form of an equation.

-4 = -3(2) + b Substitute the coordinates and slope. -4 = -6 + b +6 +6 2 = bSo our equation is y = -3x + 2, which is the equation of the line.

VIDEO LINK: Khan Academy Equations of Parallel and Perpendicular Lines

Exercises Unit 5 Section 2 Set A 1. What does it mean for two lines to be parallel?2. What is the symbol we use for parallel lines? 3. What does it mean for two lines to be perpendicular? 4. What is the symbol we use for perpendicular lines?

Find the slope of each of the given lines then draw the asked for line through the given point on a set of coordinate axes.

5. a parallel line through (4, 2) 6. a parallel line through (-2, 3)

7. a parallel line through (-1, -2) 8. a || line through (1, 5)

9. a perpendicular line through (0, 1) 10. a perpendicular line through (3, 0)

11. a line through (1, 2) 12. a line through (0, -1)

Decide from the two listed equations if the lines are parallel, perpendicular or neither.13. y = 4x + 2 14. y = x + 3 15. y = 6x + 1 y = 4x - 1 y = 2x + 3 y = -6x + 1

16. y = -4x + 2 17. y = - x + 3 18. y = x + 1

y = x - 1 y = 2x + 3 y = x - 1

19. y = - x + 2 20. y = x + 3 21. y = -x + 1

y = x - 1 y = x - 7 y = x - 8

22. y = 2 23. y = 4 24. y = 5x + 1 y = - x = 3 y = -5x - 1

25. 2y - 4x = 2 26. y + 2x = 5 27. 2y - 5x = 2 y = - x - 1 y = 2x + 3 5y = 2x - 15

28. Find the equation of the line parallel to y = x + 3 and passing through (4, -5)29. Find the equation of the line parallel to y = 3x - 2 and passing through (-1, 7)

30. Find the equation of the line to y = x - 1 and passing through (2, -1)

31. Find the equation of the line to y = - x + 3 and passing through (-2, 3)32. Find the equation of the line in slope-intercept form going through (4, -1) & (-2, -4)

Graph each pair of lines and find the point of intersection.

33. y = 2x - 5 34. y = x + 4 35. y = x - 1 y = x - 6 y = -x - 2 y = x + 1

Isolate 'y' as needed in the equations below before graphing the systems

36. 2y = -4x + 8 37. y - x = 3 38. y = -2 x + y = 1 y = x + 1 x = 0

Exercises Unit 5 Section 2 Set B1. The point where two lines intersect will make the two equations ______________.

Check to see which if the listed point will make both equations true. Answer yes or no and show your work.

2. y = -4x + 5 3. y = -.5x + 4 y = 2x - 7 y = 3x - 7

(2, -3) (2, 0)

Graph each pair of lines and find the point of intersection. 4. y = x - 4 5. y = - x - 4 y = -x + 5 y = 2x + 6

5. What does it mean for two lines to be parallel? ( List our definition. )6. Two lines will be parallel when their slopes are ___________________.7. What is the symbol we use for parallel lines? 8. What does it mean for two lines to be perpendicular? 9. Two lines will be perpendicular when their slopes are ________________________.10. What is the symbol we use for perpendicular lines?

Find the slope of the given line then draw the asked for line on a set of coordinate axes.

11. a parallel line through (-1, -3) 12. a parallel line through (-1, 4)

13. a parallel line through (1, 4) 14. a line through (-1, 2)

15. a perpendicular line through (2, -1) 16. a line through (2, 4)

Decide from the two listed equations if the lines are parallel, perpendicular or neither.17. y = 4x + 2 18. y = x + 3 19. y = -.5x + 1 y = -4x - 1 y = -3x + 3 y = -.5x + 4

20. y = 4x + 2 21. y = - x + 3 22. y = - x + 5

y = x - 1 y = x + 3 y = - x - 1

23. y = x + 2 24. y = 3 25. y = 5

y = - x - 1 y = -1 x = 5

26. y = 2 27. y = 2x + 1 ( Think carefully about problem 27 ) y = - y = 2x + 1

28. Find the equation of the line parallel to y = - x + 3 and passing through (4, 0)29. Find the equation of the line parallel to y = 4x - 2 and passing through (-2, 5)

30. Find the equation of the line to y = x - 1 and passing through (2, 1)

31. Find the equation of the line to y = -x + 3 and passing through (-5, 8)

32. Find the equation of the line in point-slope form passing through (4, 7) and (-3, 5)

Objectives The student will solve systems with an isolated variable by

substitution.

In Unit 2 we learned how to solve equations with one variable in them. The equationswe have been using in the last several units all have focused on equations with two variables, ‘x’ and ‘y’. We can solve a system of equations by making equations withtwo variables into an equation with one variable. This method for finding the solution or point of intersection is called substitution. The steps we can use for this processare listed below.

Step 1. Find the equation with an isolated variable.Step 2. Do the substitution for the isolated variable in the other

equation.Step 3. Solve the resulting equation.Step 4. Substitute the first coordinate into one of the equations

and find the second coordinate.

We will use these steps to solve the systems in the examples that follow.

Example A Given the system of equations below find the point of intersection.y = 5x – 12

2x + y = 16

Step 1. y = 5x – 12 The y is isolated in this equation.

Unit 5 Section 3

Step 2. y = 5x – 12 We take what ‘y’ is equal to and substitute it for ‘y’

2x + y = 16 2x + (5x – 12) = 16 This is the result of the substitution.

Step 3. 2x + (5x – 12) = 16 Solving the remaining equation. 2x + 5x – 12 = 16

7x – 12 = 16 +12 +12 7x = 28 7 7 x = 4

Step 4. y = 5x – 12 Substitute to find the second coordinate.

y = 5(4) – 12 y = 8

The solution is (4, 8) for this system of equations.

Example B Given the system of equations below find the point of intersection.

-x + 2y = -7 x = 3y + 2

Step 1. x = 3y + 2 The x is isolated in this equation.Step 2. x = 3y + 2 We take what ‘x’ is equal to and

substitute it for ‘x’-x + 2y = -7

- (3y + 2) + 2y = -7 This is the result of the substitution.

Step 3. - (3y + 2) + 2y = -7 Solving the remaining equation.

-3y - 2 + 2y = -7 Use the Distributive Property.

-1y - 2 = -7 + 2 +2 -1y = -5 -1 -1 y = 5

Step 4. x= 3y + 2 Substitute to find the second coordinate.

x = 3(5) + 2 x = 17

The solution is (17, 5) for this system of equations.

Example C Given the system of equations below find the point of intersection.

x = 4y - 1 2x - 8y = -2

Step 1. x = 4y – 1 The x is isolated in this equation.Step 2. x = 4y – 1 We take what ‘x’ is equal to and

substitute it for ‘x’ 2x – 8y = -2 2(4y - 1) – 8y = -2 This is the result of the substitution.

Step 3. 2(4y - 1) – 8y = -2 Solving the remaining equation.

8y - 2 – 8y = -2 Use the Distributive Property. -2 = -2

When we get a true equation that no longer has a variable this means the

two equations represent the same line! When this happens we say the lines are coincident. The two lines actually intersect at all points and

so theycoincide completely. ( “Coincidentally” in the English language means

things happen at the same time – in math it can mean they happen at the

same place. )

Example E Given the system of equations below find the point of intersection.y = 2x - 4

4x - 2y = 9

Step 1. y = 2x - 4 The ‘y’ is isolated in this equation.Step 2. y = 2x - 4 We take what ‘y’ is equal to and

substitute it for ‘y’ 4x – 2y = 9 4x – 2(2x – 4) = 9 This is the result of the substitution.

Step 3. 4x – 2(2x – 4) = 9 Solving the remaining equation. 4x – 4x + 8 = 9 Use the Distributive Property.

-8 = 9

When we get a false equation that no longer has a variable this means the two equations represent lines that are parallel! The false equations means

the lines never intersect! The only way lines can fail to intersect is if they are

parallel.

VIDEO LINK: Khan Academy Solving Linear Systems by Substitution

Exercises Unit 5 Section 3

Solve the listed systems using substitution.

1. x = y + 3 2. -x + y = 8 3. y = x + 4 3x + 2y = 14 x = 2y + 7 -2x + 3y = 10

4. 5x – 2y = 20 5. 3x + 7y = 78 6. y = 7 – 6x y = -2x + 8 y = x + 4 8x - y = 0

7. 4x – 3y = -1 8. y = 6x - 4 9. y = 4x - 5 y = -2x 6 = 9x + y y = 0

10. What does it mean for two lines to be parallel?

11. What is the symbol we use for parallel lines?

12. What does it mean for two lines to be perpendicular?

13. What is the symbol we use for perpendicular lines? _________

Find the slope of the line in each graph then draw the asked for line on another set of x-y axes.14. a parallel line through (2, 0) 15. a || line through (-2, 1)

16. a perpendicular line through (2, 3) 17. a line through (-1, 3)

Decide from the two listed equations if the lines are parallel, perpendicular or neither.

18. y = 4x + 2 19. y = x + 3 20. y = 7x + 1 y = -4x - 1 y = -2x + 3 y = 7x -4

21. y = -4x + 2 22. y = x + 3 23. y = - x + 1

y = x - 1 y = x + 7 y = x - 1

24. y = - x + 2 25. y = x - 5 26. y = x + 1

y = - x - 1 y = x + 7 y = -x - 9

27. y = 5 28. y = 0 29. y = 5x + 1 y = -6 x = 2 y = 5x + 1

30. Find the equation of the line parallel to y = 3x + 1 and passing through (2, -5)

31. Find the equation of the line parallel to y = - x - 2 and passing through (7, 1)

32. Find the equation of the line perpendicular to y = x + 4 passing through (6, -1)

33. Find the equation of the line perpendicular to y = x + 3 passing through (4, 5)

34. Given the system of linear equations

y = 4x + 5

3x – 5y = 9

a. Solve the system of equations using substitution. Write your answer as an ordered pair. Show your work algebraically and explain how you got the solution.

b. Verify that your solution is correct. Show your work algebraically.

35. Given the system of linear equations

y = x - 2 2y – 3x = -1

a. Solve the system of equations using substitution. Write your answer as an ordered pair. Show your work algebraically and explain how you got the solution.

b. Verify that your solution is correct. Show your work algebraically.

Objective The student will solve systems by isolating a variable and using

the method of substitution.

In this section we are going to be solving systems of equations by substitution again. The process will be slightly different because we will need to isolate a variable in oneof our equations before we can go through the rest of the steps. The algorithm or setof steps we should follow is given below.

Step 1. Isolate a variable in one of the equations.Step 2. Do the substitution for the isolated variable in the other

equation.Step 3. Solve the resulting equation.Step 4. Substitute the first coordinate into one of the equations

and find the second coordinate.

Unit 5 Section 4

The examples that follow show us how the process changes.

Example A Given the system of equations below find the point of intersection.

-3x + y = 7 2x + 4y = 14

Step 1. -3x + y = 7 Since ‘y’ is positive and has a coefficient of 1

+3x +3x it is easiest to isolate ‘y’ in this equation.

y = 3x + 7

Step 2. y = 3x + 7 We take what ‘y’ is equal to and substitute it for ‘y’

2x + 4y = 14 2x + 4(3x + 7) = 14 This is the result of the

substitution.Step 3. 2x + 4(3x + 7) = 14 Solving the remaining

equation. 2x + 12x + 28 = 14

14x + 28 = 14 - 28 - 28 14x = -14 14 14 x = -1

Step 4. y = 3x + 7 Substitute into the equation in which y = 3(-1) + 7 isolated the variable to find the second

y = 4The solution is (-1, 4) for this system of equations.

The overall process has not changed. We merely need to isolate a variable in one of the equations before we can do our substitution.

Exercises Unit 5 Section 4Solve the listed systems using substitution.

1. 3x + 5y = 20 2. -2x + y = 7 3. 5x - 4y = 15 x = y + 4 x = y - 3 2y = 4x - 6

4. x + y = 11 5. 3x + 4y = 2 6. 2x - y = 12 x - y = 13 x – 2y = -1 x + 4y = 51

7. 4x + 6y = -7 8. 3x + 6y = 2 9. x = y2

2x - 2y = 4 6x – 7y = 4 x + 2y2 = 75

10. The point where two lines intersect will make the two equations ______________.

Check to see which if the listed point will make both equations true.11. y = 2x - 4 12. y = -x + 5 y = -3x + 11 2y + x = 8

(3, 2) (2, 3)

Graph each pair of lines and find the point of intersection. 13. y = 3x - 4 14. y = - x + 2

y = -x + 4 y = 4

Find the slope of the line in each graph then draw the asked for line on another set of axes.

15. a || line through (-2, 2) 16. a || line through (2, 4)

17. a line through (1, 3) 18. a line through (-2, 6)

Decide from the two listed equations if the lines are parallel, perpendicular or neither.

19. y = x + 2 20. y = - x -5 21. y = 7x + 1

y = -3x - 1 y = 2x + 3 y = x - 4

22. y = 5x + 2 23. y = x + 3 24. y = - x + 1

y = -5x - 1 y = x + 7 y = -3x - 5

25. y = 1.5x + 1 26. y = x + 5 27. y = -3x + 1 y = -1.5x - 1 y = x - 9 3y = x - 6

28. y = 2 29. y = -3 30. 2y = 4x - 8 x = 2 y = 6 y + x = 1

31. Find the equation of the line parallel to y = x - 1 and passing through (4, -3)

32. Find the equation of the line perpendicular to y = - x + 4 passing through (-1, 4)33. Given the system of linear equations

2y - 4x = 14 -2y + 5x = -12

a. Solve the system of equations using substitution. Write your answer as an

ordered pair. Show your work and explain how you got the solution.

b. Verify that your solution is correct. Show your work algebraically.

Objective The student will solve systems using elimination with similar

coefficients.We have two methods to solve systems of equations. We can solve them graphically

Unit 5 Section 5

or by substitution. Both of these methods have advantages and disadvantages. Thereis, however, one more method to solve systems that we will learn. This method is called elimination or linear combination. Elimination is another way of changing equationswith two variables into a single equation with one variable. The algorithm or set of steps follow.

Step 1. Determine which variable can be eliminated by adding or subtracting.

Step 2. Add or subtract the two equations using our Properties of Equality.

Step 3. Solve the resulting equation.Step 4. Find the other coordinate of our solution.

Below are some examples of using this algorithm.

Example A Given the system of equations below find the point of intersection or solution to the system using the method of elimination.

3x + 4y = -18 3x + y = 0

When we examine the two equations we can see that the ‘x’ variable in both equations has a coefficient of 3. If we subtract 3x – 3x we get 0x or 0 and the ‘x’ terms will be gone. So we use the

Subtraction Property of Equality and subtract the two equations. ( The left and right sides of an equation are equal so we are subtracting two equal quantities. )

3x + 4y = -18 -(3x + y = 0)

3y = -18 3 3

y = -6 We have our first coordinate. 3x + y = 0 We now substitute the y coordinate 3x – 6 = 0 into one of the equations and solve.

+6 +6 3x = 6 3 3 x = 2

So the point (2, -6) is our solution or point of intersection.

Example B Given the system of equations below find the point of intersection or

solution to the system using the method of elimination. 5x + 2y = 19 2x – 2y = 9

When we examine the two equations we can see that the ‘y’ variable

in the equations has a coefficient of 2 and -2. If we -2y + 2y we

get 0y or 0 and the ‘y’ terms will be gone. So we use the Addition

Property of Equality and add the two equations. 5x + 2y = 19 2x – 2y = 9 5x + 2y = 19

+(2x – 2y = 9)7x = 28

7 7 x = 4 We have our first coordinate.

5x + 2y = 19 We now substitute the x coordinate

5(4) + 2y = 19 into one of the equations 20 + 2y = 19 and solve.

-20 -20 2y = -1 2 2

y = -

So the point (4, - ) is our solution or point of intersection.

VIDEO LINK: Khan Academy Solving Systems of Equations by Elimination

Exercises Unit 5 Section 5Solve the following by linear combination or elimination. (add or subtract the equations)Show your work. Check your answers.

1. 4x + 2y = 12 2. 3x - 4y = -11 x - 2y = 13 3x + y = -1

3. 3x + 5y = 7 4. 7x - 4y = 37 2x + 5y = 3 -2x + 4y = -12

5. 2x + 3y = 20 6. 5x – 4y = 5 2x – 4y = 6 6x – 4y = -8

7. x + 2y = 3.5 8. 3x – 7y = -24

-x + 4y = -2 2x + 7y = -16

9. x - 2y = -5 10. .5x + .8y = 5 3x - 2y = 10 .2x – .8y = -3.6

Use substitution to solve the following systems. SHOW YOUR WORK!11. y = 3x + 7 12. y = 2x - 3 y = 4x + 2 y - x = -6

13. y - 5x = 3 14. y = x + 3 3y - 2x = 22 y + 2x = -9

15. Given the circle graph below that represents the breakdown of participation in activities at Washington High School. Answer the questions that follow.

Washington High Extracurricular Activities Participation

If there are 420 students participating in Extracurricular Activities at WHS answer the questions that follow.

a. How many students were involved in fall sports?b. How many students were involved in all sports?c. How many more students were in Fall sports than in Winter sports?

16. Given the system of linear equations

2y - 5x = 13 2y + 7x = 1

a. Solve the system of equations using elimination. Write your answer as an ordered pair. Show your work and explain how you got the solution.

b. Verify that your solution is correct. Show your work algebraically.

25%

20%

Spring Sports

Other

Winter Sports Band

8%11%

34%

12%

Fall Sports

Choir

Unit 5 Section 6

Objective The student will solve systems using elimination using factors

to produce like coefficients.

When the coefficients of a system are not similar we must multiply one or both of the equations by factors that will allow us to cancel out a variable. The process is listedbelow as a set of steps or algorithm.

Step 1. Multiply one or both of the equations by factors to produce opposite coefficients and cancel out

variable terms.Step 2. Add the two equations using our Property of

Equality.Step 3. Solve the resulting equation.Step 4. Find the other coordinate of our solution.

Example A Solve the system below using elimination or linear combination.3x + 5y = -1 x - 3y = -5

Step 1. Adding or subtracting the equations now will not cancel out

out a variable’s terms. However if we multiply the bottom

equation by -3 then the ‘x’ terms will add up to 0. So we

use the Multiplicative Property of Equality and do the multiplication.

-3(x – 3y) = -5(-3) Use the Multiplication Property of Equality

-3x + 9y = 15 Use the Distributive Property

Step 2. Add the equations

3x + 5y = -1-3x + 9y = 15

Step 3. 14y = 14 Solve the resulting equation. 14 14

y = 1

Step 4. x – 3y = -5 Find the value of the x-coordinate.

x – 3(1) = -5 x – 3 = -5

+3 +3 x = -2

So our point of intersection is (-2, 1).Example b Solve the system below using elimination or linear combination.

4x – 2y = 225x + 3y = 0

Step 1. Adding or subtracting the equations now will not cancel out

out a variable’s terms. However if we multiply the bottom equation by 2 and the top equation by 3 then the ‘y’ terms will add up to 0. So we use the Multiplicative Property of Equality and do the multiplication.

3(4x – 2y) = 22(3) Use the Multiplication Property of Equality

12x – 6y = 66 Use the Distributive Property

2(5x + 3y) = 0(2) Use the Multiplication Property of Equality

10x + 6y = 0 Use the Distributive Property

Step 2. Add the equations 12x – 6y = 66 10x + 6y = 0

Step 3. 22x = 66 Solve the resulting equation. 22 22

x = 3

Step 4. 4x – 2y = 22 Find the value of the y-coordinate. 4(3) – 2y = 22

12 – 2y = 22 -12 -12 -2y = 10

-2 -2 y = -5

So our point of intersection is (3, -5).

Exercises Unit 5 Section 5Solve the following by linear combination or elimination. (add or subtract the equations)Show your work. Check your answers.

1. 3x + 5y = 1 2. 3x - 4y = -21 x - 6y = 8 8x + y = 14

3. x + 9y = 38 4. 4x – 7y = -6 2x – 5y = 7 -2x + 4y = 3

5. 2x + 3y = 20 6. 4x – 2y = -4 8x – 6y = 80 8x – 4y = -8

7. 9x + 2y = -4 8. 5x - 11y = -2 -4x + 3y = 29 2x + 8y = 24

9. x - 2y = 11 10. .8x + .7y = 11 3x - 6y = 2 .2x - .3y = -2

11. -3x - 2y = -31 12. x + 7y = 24 3x - y = 25 x - 9y = -40

Solve the following by substitution. Show your work.13. y = 3x 14. 2x – 7y = 11 y + 2x = -20 y – x = -8

List the name of the property illustrated in each problem.15. -x = -1 . x 16. a + 0 = a 17. x . y = y . x 18. 1 . z = z19. -w + w = 0 20. b . = 1 21. (b + c) + t = b + (c + t) 22. x(y – z) = xy – xz

23. Jake’s algebraic work and solution for the system of equations is shown below.

3x + 4y = -1 y + 2x = 6

Step 1 y + 2x = 6 y = 2x + 6Step 2 3x + 4y = -1 3x + 4(2x + 6) = -1Step 3 3x + 8x + 6 = -1Step 4 10x + 6 = -1Step 5 10x = 5

Step 6 x = .5Step 7 y = 2(.5) + 6 y = 7

Step 8 (.5, 7)

a. Explain what Jake did wrong in at least three of the steps.b. Check Jake’s solution algebraically and explain what the result tells

us about Jake’s solution.

Objective The student will graph linear inequalities in two variables.

In Unit 2 we learned to solve and graph inequalities in one variable. In this section we will solve and graph inequalities in two variables. The graphing process is very similarto graphing a linear equation. We will need to isolate the ‘y’ variable and use the slope and intercept to graph the line then determine the part of the graph that needs to be shaded. The algorithm for this process follows.

Step 1. Isolate ‘y’ if needed.Step 2. Find m and b.Step 3. Graph the line and draw it either solid or dashed.Step 4. Pick a point and test this point in the inequality.Step 5. Shade the side of the line that makes the inequality

true.The examples below show us how to use the algorithm.

Example A Graph y > 2x – 3

In this example the ‘y’ variable is initially isolated, so we don’t need to

perform Step 1.

y > 2x -3 m = 2 and b = -3 Step 2. Find the slope and intercept.

Step 3. We graph the line.

The line should be solid because the points on the line will make the

inequality true. We know the points on the line will make the inequality true because there is an “equal to” bar underneath theinequality symbol. For instance the point (2, 1) is on the line.We can prove this by substituting the point into the inequality.

y > 2x -3

Unit 5 Section 7

1 > 2(2) – 31 > 1 Which is true.

Example A (continued)

Now that we have graphed the line we need to pick a point. The point (0, 0) is easy to use because arithmetic with a zero is simple.

The point we will pick is (0, 0)

We will now substitute the point (0, 0) back into the inequality.y > 2x – 3 Step 4. 0 > 2(0) – 30 > -3 This is true so we should shade

the side of the line that has (0, 0) in it.

Example B Graph -2y + x > 4

There are two differences between example A and example B. The ‘y’ variable is not isolated here and the “equal to” bar is missing so we are dealing with a strict inequality.

-2y + x > 4 Step 1. Solve the inequality for y. – x – x -2y > -x + 4 -2 -2 -2 When we divide by a negative value

weswitch the inequality symbol.

Decide which side of the line toshade.

y < x – 2

Example B (continued)

y < x – 2

m = and b = -2 Step 2. Find the slope and intercept.

Step 3. We graph the line.

The line should be dashed or dotted because the points on the line willNOT make the inequality true. We know the points on the line will

make the inequality false because the points are on the line and will make the two sides of the inequality the same. But we don’t have an “equal to bar”. For instance the point (4, 0) is on the line. We can prove this by substituting the point into the inequality.

y < x – 2

0 < (4) – 20 < 2 – 2 0 < 0 Which is false!

This is why we need to have a dashed line – points on the line do notmake the inequality true. We will now substitute the point (0, 0) back into the inequality.

y < x – 2 Step 4.

0 < (0) – 20 < -2 This is false so we should shade the

side of the line that does NOT have (0, 0) in it.

Decide which side of the line toshade.

There are two important ideas we must remember:1. When to use a solid line and when to use a dashed or dotted line. 2. How to determine which side of the line to shade

VIDEO LINK: Khan Academy Graphing Linear Inequalities in Two Variables

Exercises Unit 5 Section 71. What are the steps we use to graph a linear inequality?

Graph the following inequalities 2. y < 2x - 5 3. y > -x + 7 4. y < x - 2

5. y > 3 6. x > 1 7. y > x – 5

8. y < 4 – 2x 9. y > - x 10. y < + 1

Isolate y and then graph. 11. 5y < -15x + 10 12. y - 3 > 2x - 9 13. -2(x – 6) > 2y + 8

15. -2y < x - 6 16. x – y > 5x - 2 17. 3(x + 4) + x > -y +14

Graph each pair of lines and find the point of intersection.

18. y = -2x + 4 19. y = x – 7 y = x + 7 y = 3

If we have a strict inequality which is > or < the line is dashed or dotted.If the inequality is > or < the line is solid.

To decide which side to shade we pick an easy point to use, substitute the ‘x’ and ‘y’ values into the inequality. If the inequality is true shade the side that the testpoint is on. If the inequality is false then shade the other side that the

20. What are the advantages and disadvantages of using each of the three methods we have; graphing, substitution, and elimination.

Objective The student will graph systems of linear inequalities in two

variables.

Previously in this unit we learned to solve systems of equations by graphing. We can also graphs systems of inequalities as well. The two processes are very similar. To solve a system of equations we can graph both equations and look to find the point of intersection. When we solve a system of inequalities we need to graph both of the lines in the inequalities but we need to find the region to shade that makes both of the inequalities true. The algorithm for this process is given below.

Step 1. Graph the two lines of inequalities using the technique in the previous section.

Step 2. Select a test point from each of the four regions.Step 3. Test each point in both inequalities until we find a point

that makes both inequalities true.

Step 4. Shade the region that makes both inequalities true.

We will use these steps in the examples that follow.

Example A Given the system of inequalities below, graph the solution set.

y > 4x – 3y < -x + 2

Using the slopes and intercepts we graph the lines that help form the

inequalities.

for y > 4x – 3 m = 4 and b = -3for y < -x + 2 m = -1 and b = 2

The graph of the lines would be

Unit 5 Section 8

Step 1. Graphing the two lines of the inequalities is now complete!

There are four regions into which the lines divide the x-y plane. We must

choose a point from each of the four regions.

Now we need to test each of the points in the original inequalities to see which region we need to shade.

Step 3. Testing each point in the inequalities.

Only one of the test points made both inequalities true. The point (0,0)

made both inequalities true. This is the region we will shade.

The first line must be dashed or dotted because it was a strict inequality. The second line is solid because it had the “equal to bar”.

(0, 4)

(0, -6)

(5, 0)(0, 0)

Testing (0, 4) y > 4x – 3 4 > 4(0) – 3 4 > 0 – 3 4 > -3 true y < -x + 2 4 < -(0) + 2 4 < 2 false

Testing (5, 0) y > 4x – 3 0 > 4(5) – 3 0 > 20 – 3 0 > 17 false y < -x + 2 0 < -(5) + 2 0 < -3 false

Testing (0, -6) y > 4x – 3 -6 > 4(0) – 3 -6 > 0 – 3 -6 > -3 false y < -x + 2 -6 < -(0) + 2 -6 < 2 true

Testing (0, 0) y > 4x – 3 0 > 4(0) – 3 0 > 0 – 3 0 > -3 true y < -x + 2 0 < -(0) + 2 0 < 2 true

Step 2. Picking a point from

each region. We pick points with zero’s to make

the

Step 4. Shade the appropriate region.

Example B Given the system inequalities below graph the solution set.

-3y > 2x – 6 2y + 4 < 6x + 4

In this example we must first isolate the y variables so that we can use

the slope and intercept to graph the line. 2y + 4 < 6x + 4 – 4 – 4 2y < 6x 2 2 y < 3x

Using the slopes and intercepts we graph the lines that help form the

inequalities.for y < 3x m = 3 and b = 0for y < - x + 2 m = - and b = 2

The graph of the lines would be

Step 1. Graphing the two lines of the inequalities is now complete!

Again one line must be dashed and the other solid. This does not always have to happen but it can.

-3y > 2x – 6-3 -3 -3 y < - x + 2

Since we divided by a negative value we had to switch the symbol.

There are four regions into which the lines divide the x-y plane. We must

choose a point from each of the four regions.

Now we need to test each of the points in the original inequalities to see which region we need to shade.

Step 3. Testing each point in the inequalities. ( We can use the inequalities in which we isolated y. )

Only one of the test points made both inequalities true. The point (0,-3)

made both inequalities true. This is the region we will shade.

Step 4. Shade the appropriate region.

Testing (0, 4) y < 3x 4 < 3(0) 4 < 0 false

y < - x + 2

4 < - (0) + 2 4 < 0 + 2 4 < 2 false

Step 2. Picking a point from

each region. We couldn’t pick

(0,0) since it is on one

Testing (6, 0) y < 3x 0 < 3(6) 0 < 24 true

y < - x + 2

0 < - (6) + 2 0 < -4 + 2 0 < -2 false

Testing (0, -3) y < 3x -3 < 3(0) -3 < 0 true

y < - x + 2

-3 < - (0) + 2 -3 < 0 + 2 -3 < 2 true

Testing (-1, 0) y < 3x 0 < 3(-1) 0 < -3 false

y < - x + 2

0 < - (-1) + 2

0 < + 2

0 < 2

true

(-1, 0)

(0, -3)

(6, 0)

(0, 4)

VIDEO LINK: Khan Academy Graphing Systems of Inequalities in 2 Variables

Exercises Unit 5 Section 81. List the algorithm we have for solving a systems of inequalities.

Graph the systems of inequalities

2. y > x – 3 3. y > 4x – 6 4. y > 3x – 1 y < 2x + 1 y < -2x + 1 x > 1

5. y < -x + 3 6. -2y < 6x – 6 7. y > 3x – 1 y < 3x - 5 y < x + 1 y < -1

8 y > 1 9. y < 4x 10. y > x + 1 x < 0 2y < -3x – 4 y < -3

y < -x + 2

Graph the following inequalities 11. y < x - 5 12. -y < 2x - 4 13. y + 1 > x - 3

14. Find the slope of the line that passes through the points (2,4) and (-1, 6).

15. Find the equation of the line in slope-intercept form that has a slope of m = 3 and passing through the point (-4, -5).

16. Find the equation of a line in point-slope form that has a slope of m = -2 and passing through the point (-2, 7).

17. Find the equation of a line in slope-intercept form that passes through the points (-3, 4) and (0, 10).

18. Find the equation of the line in slope-intercept form that is parallel to y = 5x + 1 and passing through the point (2, -4).

19. If a line has a slope of m = find the slope of a perpendicular line.

20. Convert the equation y = x – 1 to standard form.

21. Mark’s algebraic work and solution for the system of equations is shown below.

5x + 2y = -2 -2x + 3y = 14

Step 1 5x + 2y = -2 3(5x + 2y = -2) 15x + 6y = -6

Step 2 -2x + 3y = 14 -2(-2x + 3y = 14) -4x - 6y = -28

Step 3 15x + 6y = -6 -4x - 6y = -28 12x = -36

Step 4 x = -3Step 5 -2x + 3y = 14

-2x + 3(-3) = 14 Step 6 -2x - 6 = 14

Step 7 -2x = 20Step 8 x = -10

(-3, -10)

a. Explain what Mark did wrong in at least three of the steps.b. Check Mark’s solution algebraically and explain what the result tells

us about Mark’s solution.

Translate each sentence into an inequality.

22. Twice a number ‘x’ increased by six is at least ninety.

23. Half a number ‘a’ decreased by seven is at most thirteen.

24. Jane has sold more than four times as many raffle tickets as Sally.

25. Alex’s high score in the video game “Angry Kangaroos” is less than one-third of martin’s score.

Objective The student will solve contextual problems with systems of

equations and inequalities.

Systems always have two equations or inequalities. When we deal with contextual problems our first task will be write two equations or inequalities. Some steps that may be useful in creating these equations or inequalities are given below.

Step 1. List what the variables will represent.Step 2. List the quantities or constants we see in the problem.Step 3. Determine how to calculate the quantities.Step 4. Write the two equations or inequalities.Step 5. Solve the system.

The applications for systems in this section can be broken into several categories. The first category we will deal with involves translation. These problems use our knowledge of operations and the words that represent them. These problems may not require that we use all of the steps above. Some examples of this type follow.

Example A The sum of two numbers is 18 and the difference of the same two numbers is 14. Find the numbers.

Step 1. The problem says to find two numbers, and in all of these

exercises we will need to have two variables - one for each number.

x = first number y = second number

Step 2. The constants in the problem are 18 and 14. In this problem they are simply numbers.

Step 3. The key words are “sum” and “difference” which means one equation will need to use addition and the other subtraction.

Step 4. The two equations are:

Unit 5 Section 9

x + y = 18 The equation for the sum.x – y = 14 The equation for the difference.

Step 5. Solving the system. We can use any of three methods.

Graphing may not be practical because our numbers could be large or they might not work out evenly. (With a graphing calculator this method could be a good

alternative.)

When we look at the system below we can see that adding the two equations will cause the ‘y’ variable to cancel out.We should use the method of elimination.

x + y = 18x – y = 14 2x = 32 2 2 x = 16

x + y = 18 Now we find the other number. 16 + y = 18 -16 -16

y = 2

Our two numbers are 16 and 2.

Example B The sum of two numbers is 74. The larger number is eight morethan twice the smaller. Find the numbers.

Step 1. The problem says to find two numbers, and in all of these

exercises we will need to have two variables - one for each number.

x = smaller number y = larger number

Step 2. The constant in the problem is 74.

Step 3. The key word is “sum” one equation will need to use addition. The other equation can be found by direct

translation of the second sentence in the problem.

Step 4. The two equations are:

x + y = 74 The equation for the sum.y = 2x + 8 The direct translation

Step 5. Solving the system. We can choose any of three methods.

When we look at the system below we can see that in the bottom equation the ‘y’ is isolated. We can use the method of substitution.

x + y = 74y = 2x + 8

x + 2x + 8 = 74

3x + 8 = 74 -8 -8

3x = 66 3 3

x = 22

x + y = 74 Now we find the other number.

22 + y = 74 -22 -22

y = 52

Our two numbers are 22 and 52.

Another category of problems could be called the “together” type of problem. We follow the same set of steps or algorithm. The example below shows us how to attack this type of problem.

Example C Mike and Tom are players on the Washington High School Basketball team.

In a recent game Mike scored five less than four times as many points as

Tom. Together they scored 30 points. Find the number of each person scored.

Step 1. The problem says to find two numbers, and in all of these

exercises we will need to have two variables - one for each number.

x = Tom’s points The smaller number y = Mike’s points The larger number

Step 2. The constant in the problem is 30 points.

Step 3. The key word is “together”, so one equation we will need to

use is addition. The other equation can be found by direct translation of the second sentence in the problem.

Example C (continued)

Step 4. The two equations are:

x + y = 30 The equation for together.y = 4x - 5 The direct translation

Step 5. Solving the system. We can choose any of three methods.

When we look at the system below we can see that in the bottom equation the ‘y’ is isolated. We can use the method of substitution.

x + y = 30y = 4x - 5

x + 4x - 5 = 30

5x – 5 = 30 +5 +5

5x = 35 5 5 x = 7 This is Tom’s points.

x + y = 30 Now we find Mike’s points. 7 + y = 30 -7 -7

y = 23

So Mike scored 23 points and Tom scored 7

A third category or type of problem can be called coin problems. The example below

shows us how to use our steps for this category.

Example D Mr. Jones wants his class to solve a problem for him. He has a small glass

jar with coins in it. The jar contains only dimes and nickels. He tells his

class the jar has $5.00 in it. He also tells his students that the total

number of coins is 71. The first student to find the numbers of nickels

and the number of dimes in the jar gets to keep the coins.

Step 1. The problem says to find two numbers, and in all of these

exercises we will need to have two variables - one for each number.

x = the number of nickels y = the number of dimes

Example D (continued)

Step 2. The constants in the problem are 71 which is the number of coins and $5.00 which is the value of the coins.

Step 3. The word “total” tells us we have to add the numbers of coins together to get the 71. To find the value of the value of the coins we have to look at how we can find the value of just one type of coin. If we had 2 nickels we could multiply $0.05 times 2 and get $0.10. If we had 6 nickels we could multiply $0.05 times 6 and get $0.30. So if we have ‘x’ nickels we would multiply $0.05 times x. We would write this as .05x. We can do the same thing for the dimes and we would get .10y or .1y. These two expressions for the value must be added together to get the total value.

Step 4. We should be able to write the equations now.

x + y = 71 The total number of coins .05x + .1y = 5 The total value of the coins.

Step 5. Solving the system. We can choose any of three methods.

One method for solving this system would be to use elimination and multiply the bottom equal by -20. This will allow us to cancel out the ‘x’ terms.

x + y = 71-20(.05x + .1y) = 5(-20)

x + y = 71

-x - 2y = -100 After multiplying by -20.

x + y = 71 -x - 2y = -100

-1y = -29 -1 -1 y = 29 The number of dimes

x + y = 71 Now we find the number of nickels. x + 29 = 71 -29 -29

x = 42 the number of nickels

So there are 42 nickels and 29 dimes.Example E An insecticide must be diluted before it can be sprayed. Currently the

solution is 60% insecticide. The most efficient and safe solution is a

25% solution. The current volume of the solution is 2600 ml. What

volume of water must be added to make the solution 25%.

Step 1. This problem involves amount of a solution. We need to know

how much water needs to be added and how much solution

we will have when we are done.

x = the volume of water to be addedy = the volume of the solution after adding water

Step 2. The two quantities in the problem are the amounts of liquid

and the amount of insecticide.

Step 3. One of the equations should give us the “total” amount of

liquid. The other equation will be based on the percentages

that will help us calculate the amount of insecticide, which

must be the same before and after adding the water. Adding

water will not change the amount of insecticide in the mixture. We use percentages in this case to multiply.

Step 4. Our equations are.

y = x + 2600 The total amount of liquid .6(2600) = .25(y) The amount of insecticide

We need to solve the second equation and then substitute.

Step 5. 1560 = .25y .25 .25

6240 = y The second equation solved

y = x + 2600 Now we substitute 6240 = x + 2600 -2600 -2600 3640 = x

So our answer is we must add 3640 ml of water.

These techniques can be used on many types of problems not just the types we have listed. When we do the problems we need to show steps 1, 4, and 5. Steps 2 and 3in our algorithm must be performed mentally.

Exercises Unit 5 Section 9Solve the following problems. Show your work.

1. The sum of two numbers 51. The difference of the numbers is 9. Find the numbers.

2. Two numbers have a sum of 45 and a difference of 13 find the numbers.

3. The sum of two numbers is 26 and the difference of the same two numbers is 4. Find the numbers.

4. The sum of two numbers is -34 and the difference of the numbers is 8. Find the numbers.

5. The sum of two numbers is 61. The larger number is thirteen more than twice the smaller number. Find the numbers.

6. The difference of two numbers is 72. The larger number is sixteen more than five times the smaller number. Find the numbers.

7. A rectangle has a perimeter of 96 cm. The length is six less than twice the width. Find the length and width.

8. The width of a rectangle is half the length. If the perimeter 81 inches. Find the length and width.

9. A rectangle has a perimeter of 68 m. The length is eight meters more than the width. Find the AREA of the rectangle.

10. A rectangle has a perimeter of 56 cm. If we were to add 5 cm to the width and subtract 5cm from its length we would have a square. Find the length and width.

11. Washington High School has a fund raiser on Halloween. They are running a haunted house. The tickets are $5 for children and $10 for adults. At the end of the night they know they had $1705. They also know they had 273 people go through the haunted house. How many of each type of ticket was sold?

12. Sam is buying clothes for the start of the school year. He buys a number of pairs of pants and some shirts. The pants cost $42 each and the shirts cost $38 dollars each. The total cost was $316. If he bought 8 items of clothing altogether find how many shirts he bought.

13. The local zoo is running a fund raiser. They are holding a pancake breakfast. There is a fixed price for each breakfast but they charge extra for coffee. One table has a total bill of $84 for 6 orders of pancakes and 4 cups of coffee. A second table is charged $77.50 for 5 orders of pancakes and 5 cups of coffee. Find the price of the coffee and pancakes.

14. Alicia manages a toy store. She is buying two sizes of teddy bears, large and small. The large bears cost twice as much as the small bears. If she buys 20 small bears and 12 large bears and the cost is $448.80 find the cost of each size bear she bought.

15. A chemist has 2000 g of a solution. The solution is 30% acid. He wants to use the

solution to wash some metal equipment but the acid content is too high and will damage the metal. He needs to add water to dilute the acid. If he wants the solution to be 12% acid how much water does he need to add?

16. A chemical engineer wanted to etch glass with an acidic solution. The solution that he had a 25% acid content. This is too weak to do the etching. He wanted to raise the content to 40%. He added 800 g of acid to the solution and raised the content to 40%. What was the size of the original solution?

17. A new High School just opened. The students were asked to vote on the mascot. The choice they had was between the “Wolves” and the “Hawks”. In the voting the total number of votes cast was 924. If the “Wolves” won by 90 votes how many votes did each mascot received?