Alg 1 Sem 2 Practice Final Solutions

7/27/2019 Alg 1 Sem 2 Practice Final Solutions

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7532 CP Algebra 1 Second Semester Practice Final Solutions (Problems 1-23)

Sample Problem 1: Graph a linear inequality in two variables

Graph the solution set of the inequality: 4x y > 2

Step 1: 4x y > 2

-4x -4x

Step 2: - y > -4x + 2-1 -1 -1

Step 3: y < 21

4

x m =1

4b = -2

Step 4:

Sample Problem 2: Solve a system using the Substitution Method

Solve the system: 3x 2y = 11

y = 3x - 1

Step 1: 3x 2(3x 1) = 11

Step 2: 3x 6x + 2 = 11

-3x + 2 = 11

-2 -2

-3x = 9

-3 -3

x = -3

Step 3 y = 3(-3) 1, y = -9 1, y = -10

Step 4 Solution: (-3, -10)

Sample Problem 3: Solve a system using the Elimination method

Solve the system: 2x + 3y = 8

-5x + 2y = -1

Step 1: 5(2x + 3y = 8)

2(-5x + 2y = -1)

Step 2: 10x + 15y = 40

-10x + 4y = -219y = 38

19 19

y = 2

Step 3: 2x + 3(2) = 8

2x + 6 = 8

-6 -6

2x = 2 = 1

2 2

Step 4: Solution: (1, 2)

Step 1: Identify one equation solved for a single variable

(y = 3x 1). Replace that variable (y) in the other

equation with the expression that it is equal to (3x 1)

Step 2: Solve for the remaining variable in the equation.

Step 3: Substitute the value found for that variable

(x = -3) into the other equation and solve for the

remaining variable (y = -10)

Step 4: The solution is written (x, y) or (-3, -10)

Step 1: Multiply one or both equations by some value to

result in the same coefficient with opposite signs in front

of the same variable. Multiplying the first equation by 5

and the second equation by 2 results in the coefficients

10 and -10 in front of x.

Step 2: Add the resulting equations eliminating the

variable with same and opposite coefficients and solve

for the remaining variable.

Step 3: Substitute the value found for that variable

(y = 2) into the other equation and solve for the

remaining variable (x = 1)

Step 4: The solution is written (x, y) or (1, 2)

Step 1: Put the equation in y > mx + b form by

subtracting 4x from both sides and dividing by -1, the

coefficient of y.

Step 2: When dividing both sides by a negative, the

inequality flips.

Step 3: Reduce the fractions and identify the slope (1

4)

and the y-intercept (-2).

Step 4: Plot the y-intercept, move 4 (numerator of slope)

in the y-direction and 1 (denominator of slope) in the x-

direction and plot another point. Connect the points with

a dashed line if the inequality is less than or greater than

but not equal to. Shade according to the inequality on y

(if y >, then shade above the line, if y


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Sample Problem 4: Simplify exponent expressions (division)

Simplify :7

23

8

12

xy

yx

Step 1: =7

23

2

3

xy

yx

Step 2: =5

2

3

2

y

x

Sample Problem 5: Simplify exponent expressions (multiplication)

Simplify : )7)(3( 245 yxyx

Step 1: =3921 yx

Sample Problem 6: Simplify exponent expressions (powers)

Simplify :435 )2( yx

Step 1: =122016 yx

Sample Problem 7: Simplify exponent expressions involving negative exponents

Simplify :35

12

15

10

yx

yx

Step 1: =35

12

3

2

yx

yx

Step 2: =13

52

3

2

yy

xx

Step 3: =4

7

3

2

y

x

Sample Problem 8: Add or subtract polynomials

Simplify: )532()7412( 22 ++ xxxx

Step 1: = 5327412 22 ++ xxxx -54 -54

Step 2: = 12710 2 + xx

Step 1: Reduce the coefficients by canceling common

factors.

Step 2: Combine like terms by subtracting the exponents

on each term, subtracting the smaller exponent from the

larger exponent.

Step 1: Combine like terms by multiplying the

coefficients and adding the exponents.

Step 1: Simplify by raising the coefficient to the power

of the expression (24) and multiplying the exponents on

the variables by the power of the expression.

Step 1: When subtracting, change the sign of each term

of the polynomial that is being subtracted.

Step 2: Combine like terms

Step 1: Reduce the coefficients by canceling common

factors.

Step 2: Simplify variables with negative exponents by

moving them from the numerator to the denominator or

from the denominator to the numerator and making the

exponent positive.

Step 3: Combine like terms by adding the exponents.


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Sample Problem 9: Multiply a monomial by a polynomial

Multiply: )1752(3 232 ++ xxxx

Step 1: = 2345 321156 xxxx ++ -54 -54

Sample Problem 10: Multiply binomials

Multiply: (3x 2)( x + 6)

Step 1: = 1221832

+ xxx

Step 2: = 121632

+ xx

Sample Problem 11: Factor by taking out the greatest common factor

Factor completely: 15x2

5x

Step 1: 5x(3x 1)

Sample Problem 12: Factor trinomials in the form x2

+ bx + c

Factor Completely: x2

12x + 20

Step 1: (x - )(x - )

Step 2: (x 10)(x 2)

Sample Problem 13: Factor trinomials in the form x2

+ bx + c

Factor Completely: x2

- x - 42

Step 1: (x + )(x - )

Step 2: (x + 6)(x 7)

Sample Problem 14: Factor difference of squares


81

Step 1: (2x + )(2x - )

Step 2: (2x + 9)(2x 9)

Step 1: Divide both terms by the greatest common factor

of each term (5x). Write the expression by showing the

greatest common factor multiplied by the two remainingterms.

Step 1: Separate into two binomials with first term x. If

the last term is positive (+20) then the signs in the

binomials are the same and they have the sign of the

middle term (-12x).

Step 2: Complete the binomials with two values that

multiply together to get the last term (+20) and add to the

coefficient of the middle term (-12). The two number

are -10 and -2.

Step 1: When both terms are perfect square and are

being subtracted, separate into two binomials where the

first term is the perfect square factor of the leading term

(2x). The signs are different.

Step 2: Complete the binomials with the perfect square

factor (9) of the last term.

Step 1: Separate into two binomials with first term x. If

the last term is negative (-42) then the signs in the

binomials are different.

Step 2: Complete the binomials with two values that

multiply together to get the last term (-42) and add to the

coefficient of the middle term (-1). The two number are

-7 and +6.

Step 1: Multiply each term of the first binomial by each

term of the second.

Step 2: Combine like terms

Step 1: When multiplying, distribute the monomial

( )3 2x by multiplying the coefficient of each term and

adding the exponents on the variable for each term


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Sample Problem 15: Factor trinomials in the form ax2

+ bx + c


- 7x - 15

Step 1: 2(-15) = -30

Step 2: -10 and 3

Step 3: 2x2

10x + 3x 15

Step 4: (2x2

10x) + (3x 15)

Step 5: 2x(x 5) + 3(x - 5)

Step 6: (2x + 3)(x 5)

Sample Problem 16: Solve by Factoring

Solve for x: x2

- 3x = 54

Step 1: x2

- 3x = 54

-54 -54

x2

- 3x 54 = 0

Step 2: (x 9)(x + 6) = 0

Step 3: x 9 = 0 or x + 6 = 0

+ 9 + 9 -6 -6

x = 9 or x = -6

Step 4: Solution {-6, 9}

Sample Problem 17: Solve by Completing the Square

Solve for x: x2

+ 12x 5 = 0

Step 1: x2

+ 12x 5 = 0

+ 5 + 5

x2 + 12x = 5

Step 2: x2

+ 12x + 36 = 5 + 36

Step 3: (x + 6)2

= 41

Step 4: 2)6( +x = 41

Step 5: x + 6 = 41

- 6 -6

Step 6: x = 416

or {0.40, 12.40}

Step 1: Multiply the coefficient of the first term with the

last term. (-30)

Step 2: Find two factors of the resulting number that add

to the coefficient of the middle term (-10 + 3 = -7)

Step 3: Replace the middle term using the two factors

previously found as coefficients.

Step 4: Group the resulting first two and last two terms.

Step 5: Factor out the greatest common factor from each

pair of grouped terms.

Step 6: Create two binomials by pairing up the factored

out greatest common factors (2x and 3) into one

binomial and using the common result (x 5) as the

other binomial.

Step 1: Set the equation equal to zero by subtracting 54

from both sides.

Step 2: Factor the resulting trinomial on the left.

Step 3: Solve for x by setting each factor equal to zero

and solving.

Step 4: List the solutions in a set.

Step 1: Move the constant term to the right side of the

equation by adding 5 to both sides.

Step 2: Add a value (36) to both sides to create a perfect

square trinomial on the left side. This number is found

by taking the coefficient of the 2nd

term (12), dividing by

2, and squaring the result.

Step 3: Factor the left side.

Step 4: Take the square root of both sides

Step 5: The square root cancels the square on the left

resulting in a positive or negative possibility for the

variable. The positive or negative is transferred to the

right side.

Step 6: Get x by itself by adding 6 to both sides. The

solution can be kept in square root form or computed

into decimal form.


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Sample Problem 18: Solve by the using the Quadratic Formula

Solve for x: 2x2

+ 6x = 3

Step 1: 2x2

+ 6x = 3

-3 -3

2x2

+ 6x 3 = 0

Step 2: a = 2, b = 6, c = -3

Step 3: x =a

acbb

2

42

x =)2(2

)3)(2(4)6()6( 2

Step 4: x =4

24366 +

Step 5: x =4

606 = {0.44, -3.44}

Sample Problem 19: Graph a quadratic function using the x-intercepts

Graph the function y = x2

4x + 3

Step 1: 0 = x2

4x + 3

Step 2: 0 = (x 3)(x 1)

0 = x 3 or 0 = x 1

3 = x or 1 = x

Step 3:

Sample Problem 20: Find the number of x-intercepts of a quadratic

function using the discriminant

Determine if the graph of the function y = 3x2

+ 2x + 6

would have 2, 1 or 0 x-intercepts.

Step 1: a = 3, b = 2, c = 6

Step 2: b2

4ac = (2)2

4(3)(6)

= 4 72

= -68

Step 3: There are zero x-intercepts

Step 1: Set y = 0 to find the x-intercepts

Step 2: Solve the resulting equation by factoring,

completing the square, or by the quadratic formula.

Step 3: The solutions represent the x-intercepts of the

graph. Plot the x-intercepts on the x-axis. Graph a

sketch of the parabola through the x-intercepts. If the

coefficient of x2

is positive, then the parabola opens up.

If the coefficient of x2

is negative, then the parabola

opens down.

Step 1: Set the equation equal to zero by subtracting 3

from both sides.

Step 2: Identify the coefficients of the three terms (a =

coefficient of x2, b = coefficient of x, c = constant)

Step 3: Substitute these values into the quadratic

formula.

Step 4: Simplify by squaring and multiplying.

Step 5: Simplify by combining values under the square

root.

Step 6: The solution can be kept in square root form or

computed into decimal form.

Step 1: Identify the coefficients of the three terms (a =

coefficient of x2, b = coefficient of x, c = constant)

Step 2: Substitute these values into the discriminant

(b2 4ac) and compute the value of the discriminant.

Step 3: Determine the number of x-intercepts from the

value of the discriminant according to the following rule:

b2

4ac > 0 two x-intercepts

b2

4ac = 0 one x-intercept

b2

4ac < 0 zero x-intercepts

Therefore a discriminant value of -68 results in zero x-

intercepts.


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Sample Problem 21: Reduce a rational expression

Simplify:107

252

2

++

xx

x

Step 1: =)2)(5(

)5)(5(

++

+

xx

xx

Step 2: =2

5

+

x

x

Sample Problem 22: Multiply rational expressions

Simplify:25

246

127

15222

2

+

+

x

x

xx

xx

Step 1:)5)(5(

)4(6

)3)(4(

)5)(3(

+

+

xx

x

xx

xx

Step 2:)5)(5(

)4(6)3)(4()5)(3(

+

+

xxx

xxxx

Step 3:5

6

x

Sample Problem 23: Add or subtract rational expressions

Subtract and simplify:4

3

127

12

+

++

+

xxx

x

Step 1:4

3)4)(3(

1+

++

+

xxx

x

Step 2:)3(

)3(

)4(

3

)4)(3(

1

+

+

+

++

+

x

x

xxx

x

Step 3:)4)(3(

)3(31

++

++

xx

xx

)4)(3(

931

++

+

xx

xx

)4)(3(

82

++

xx

x

Step 4:)4)(3(

)4(2

++

+

xx

x

Step 5:3

2

+

x

Step 1: Factor the numerator and the denominator

Step 2: Cancel common factors in the numerator and

denominator.

Step 1: Factor each expression in each fraction by using

methods of trinomial factoring, difference of squares, or

taking out the greatest common factor.

Step 2: Cancel common factors in the numerator and

denominator. Common factors may also be canceled

across fractions.

Step 3: Express the remaining factors as a singlefraction.

(Note: Dividing rational expressions would involve the

same procedure since division can be changed to

multiplication by taking the reciprocal of the second

fraction (the divisor).)

Step 1: Factor the denominator to see which factors are

in the denominator of the two fractions

Step 2: Make the denominators the same by multiplying

the numerator and denominator of a fraction by any

factors that are in the denominator of the other fraction.

The second fraction is missing the factor (x + 3) so it

needs to be multiplied by that factor.

Step 3: Simplify into a single fraction by subtracting the

numerators and keeping the denominator the same.

Subtracting the numerator includes distributing the 3 and

subtracting both terms of the second fraction. Then

combine like terms.

Step 4: Factor the numerator to see if there are any

matching factors that can be canceled.

Step 5: Cancel the matching factor (x + 4).

Alg 1 Sem 2 Practice Final Solutions

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