MATH 082 FINAL EXAM REVIEW –SOLUTIONS TO SELECTED PROBLEMS
(Due to time constraints, solutions will be provided to a selected subset of the final review problems. Answers to all
problems are provided in a separate document also available at www.svsu.edu/tutor.)
1.
Determine which of the numbers in the set are rational numbers. { √10, −3⁄8 , −3 𝜋, −7.3, 5, √−27}
3
Solution: −3⁄8 , −7.3, 5, √−27
Recall that a number is rational if it can be written as a ratio of two integers. −3⁄8 is already written as a ratio of
3
3
73
two integers. −7.3 = −7 10 = − 10. (Since a terminating decimal can always be written as a mixed number or
5
1
decimal with a power of ten in the denominator, all terminating decimals are rational numbers.) 5 = . (Since any
3
3
integer can be written as itself over one, all integers are rational numbers.) √−27 = −3 = − 1.
2. Evaluate: 24 + 9 − 5(−1)3
Solution: 24 + 9 − 5(−1)3 = 16 + 9 − 5(−1) = 16 + 9 + 5 = 30
3. Identify the property of algebra illustrated by the statement.
4(𝑥𝑦) = (𝑥𝑦)4
Solution: Commutative Property of Multiplication. (This states that when two numbers are multiplied, the order of
the multiplicands doesn’t matter; switching the order gives the same answer.)
4. Simplify: 3𝑦 2 − 8(4𝑦 − 2𝑦 2 ) − 𝑦
Solution: 3𝑦 2 − 8(4𝑦 − 2𝑦 2 ) − 𝑦 = 3𝑦 2 − 32𝑦 + 16𝑦 2 − 𝑦 = 19𝑦 2 − 33𝑦
5. Translate the phrase into an algebraic expression . The product of 7w and 4z subtracted from 22.
Solution: 22 − (7𝑤)(4𝑧) (Recall that “product” means multiplication. Also be careful about the order in which you
write down the subtraction.)
6. Evaluate −𝑥 3 − 2𝑦 2 when x = 1 and y = −2
Solution: When x = 1 and y = −2, −𝑥 3 − 2𝑦 2 = −13 − 2(−2)2 = −1 − 2(4) = −1 − 8 = −9
7. Determine if z =
2
3
Solution: When z =
When z =
2
3
is a solution to 3(4 − 𝑧) = −6(−1 − 𝑧)
2
3
2
2
, 3(4 − 𝑧) = 3 (4 − 3) = 3 ∙ 4 − 3 ∙ 3 = 12 − 2 = 10
2
2
, −6(−1 − 𝑧) = −6 (−1 − 3) = (−6) ∙ (−1) − 6 ∙ (− 3) = 6 + 4 = 10
Since 10 = 10, z =
2
3
is indeed a solution to 3(4 − 𝑧) = −6(−1 − 𝑧).
8. Solve: −5𝑦 − 7 = −13𝑦 + 19
Solution: Add 13𝑦 to both sides to get −5𝑦 − 7 + 13𝑦 = −13𝑦 + 19 + 13𝑦
⇒ 8𝑦 − 7 = 19. Now add 7 to both sides to get 8𝑦 − 7 + 7 = 19 + 7
⇒ 8𝑦 = 26. Finally, divide both sides by 8 to get the solution: 𝑦 =
26
8
=
13
4
9. Solve: −3(4 + 4𝑥) = −12𝑥 − 2
Solution: Distribute the 3 first on the left side to get −3(4) − 3(4𝑥) = −12𝑥 − 2
1
⇒−12 − 12𝑥 = −12𝑥 − 2. Now add 12x to both sides to get −12 − 12𝑥 + 12𝑥 = −12𝑥 + 12𝑥 − 2
⇒ −12 = −2. This is not possible, therefore the answer is No Solution.
𝑥−6
10. Solve:
3
=
𝑥+5
−9
Solution: Use cross multiplication to get −9(𝑥 − 6) = 3(𝑥 + 5). Then distribute the −9 and the 3 to get
⇒ −9(𝑥) − 9(−6) = 3(𝑥) + 3(5) ⇒ −9𝑥 + 54 = 3𝑥 + 15. Now add 9𝑥 to both sides to get
⇒ +9𝑥 − 9𝑥 + 54 = 9𝑥 + 3𝑥 + 15 ⇒ 54 = 12𝑥 + 15. Now subtract 15 from both sides to get
⇒ 54 − 15 = 12𝑥 + 15 − 15 ⇒ 39 = 12𝑥. Now divide by both sides by 12 to get
⇒
39
12
=
12𝑥
12
⇒𝑥=
39
12
=
13
.
4
So, our solution is 𝑥 =
13
.
4
11. Solve the percent equation: 70.2 is 54% of what number?
Solution: Translate the equation remembering that is means equals and of means multiplication. So the equation is
⇒ 70.2 = .54(𝑥), where the percent is expressed as a decimal and x is the number we are trying to find.
Now divide both sides by .54 to get
70.2
.54
=
.54𝑥
.54
⇒ 130 = 𝑥. So, 70.2 is 54% of 130.
12. Solve and graph on a number line: -3x- 16 < -25
Solution: Start by adding 25 to both sides to get −3𝑥 − 16 + 25 < −25 + 25
⇒ −3𝑥 + 9 < 0. Now subtract 9 from both sides to get −3𝑥 < −9. Next, divide both sides by -3. However, make
sure to switch the inequality since you are dividing by a negative number. So 𝑥 > 3.
So graph an arrow in the positive direction with an endpoint at 𝑥 = 3. Indicate that 𝑥 = 3 is not a solution with
either an open circle or rounded parenthesis at 𝑥 = 3.
13. Determine if (-2,-8) is a solution to 𝑦 = −5𝑥 − 2.
Solution: Substitute −2 for 𝑥 and −8 for 𝑦 to get −8 = −5(−2) − 2
⇒ −8 = 10 − 2
⇒ −8 = 8 This is not true, so (-2,-8) is not a solution of 𝑦 = −5𝑥 − 2.
14. Sketch the graph of y= |𝑥 + 3|
Solution:
We can see that the expression within the absolute value sign is equal to 0 when 𝑥 = −3. Thus, when
constructing a table of values, consider values both greater than and less than -3 to get a better idea of the
behavior of the function.
x
-5
-4
-3
-2
-1
y
|−5 + 3| = |−2| = 2
|−4 + 3| = |−1| = 1
|−3 + 3| = |0| = 0
|−2 + 3| = |1| = 1
|−1 + 3| = |2| = 2
These points are plotted in the graph above. These points indicate a v-shaped function, as shown below.
2
15. Sketch the graph of −3𝑥 + 5𝑦 = −15 by first finding the x-and y- intercepts.
Solution:
To find the x-intercept, we set 𝑦 equal to 0. So we get −3𝑥 + 5(0) = −15
⇒ −3𝑥 = −15. Dividing both sides by −3, we get 𝑥 = 5. So the x-intercept is (5,0).
To find the y-intercept, set 𝑥 equal to 0. Thus, we get −3(0) + 5𝑦 = −15
⇒ 5𝑦 = −15. Dividing both sides by 5, we get 𝑦 = −3. Thus, the y-intercept is (0, -3)
These points are plotted above, with a line connecting them.
16. Sketch the graph of −3𝑥 + 2𝑦 = −6 by first writing the equation in slope-intercept form.
Solution:
To write an equation in slope-intercept form, solve it for 𝑦. First, add 3𝑥 to both sides. We get
3
2𝑦 = 3𝑥 − 6. Now, divide both sides by 2 to get 𝑦 = 2 𝑥 − 3. Slope-intercept form is 𝑦 = 𝑚𝑥 + 𝑏, where 𝑚 is
the slope of the line and 𝑏 is the y-value of the y-intercept. Thus, we know that the y-intercept is (0, -3). We
also know that the slope is
3
2
. That means that every time we move two to the right from a point on the line,
we move three up to get another point on the line. So (0+2, -3+3) = (2,0) is also a point on the line. These points
are plotted below, with a line connecting them.
3
17. Determine the slope of a line that passes through (8,-2) and (-1,5)
Solution: The formula for finding the slope of a line from two points on the line is 𝑚 =
(𝑥1 , 𝑦1 ) = (8, −2) and (𝑥2 , 𝑦2 ) = (−1,5), we get 𝑚 =
5−(−2)
−1−8
7
𝑦2 −𝑦1
.
𝑥2 −𝑥1
Substituting
7
= −9 = − 9.
Note: It does not matter which point you choose to be (𝑥1 , 𝑦1 ) and which point you choose to be (𝑥2 , 𝑦2 ).
Switching the two points will just change the sign of both the top and the bottom. These two sign changes will
cancel out to give you the same final answer.
18. Determine the slope of the line with the given equation. 7𝑥 = −2𝑦 − 11
Solution: First, write the equation in slope- intercept form by solving for 𝑦. Add 2𝑦 to both sides to get
7
7𝑥 + 2𝑦 = −11. Subtract 7𝑥 from both sides to get 2𝑦 = −7𝑥 − 11. Divide both sides by 2 to get 𝑦 = − 2 𝑥 −
11
2
7
Since slope-intercept form is = 𝑚𝑥 + 𝑏 , we know that the slope is given by 𝑚 = − 2 .
19. Write the standard form of the line through (6, -3) and has slope m=
−3
5
3
Solution: Use the point-slope form of a line: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) with (𝑥1 , 𝑦1 ) = (6, −3) and 𝑚 = − 5:
3
𝑦 + 3 = − 5 (𝑥 − 6) . We need to put this equation into standard form; i.e. we need to put it in the form 𝑎𝑥 + 𝑏𝑦 +
𝑐 = 0. First multiply both sides by 5 to clear denominators: 5(𝑦 + 3) = −3(𝑥 − 6) ⇒
5𝑦 + 15 = −3𝑥 + 18 ⇒ 3𝑥 + 5𝑦 + 15 = 18 ⇒ 3𝑥 + 5𝑦 − 3 = 0 . Thus the standard form of the desired equation
is 3𝑥 + 5𝑦 − 3 = 0.
20. Sketch the graph of the inequality 3𝑥 − 3𝑦 − 9 < 0
4
Solution: First we graph the line 3𝑥 − 3𝑦 − 9 = 0 ⇒ 3𝑥 − 3𝑦 = 9. There are many ways to graph this
equation, but here we will graph it by plotting the intercepts. To find the x-intercept, set 𝑦 = 0, so we have
3𝑥 − 3(0) = 9 ⇒ 3𝑥 = 9. Divide both sides by 3 to get 𝑥 = 3. Thus, the x-intercept is (3, 0).
To find the y-intercept, set 𝑥 = 0 so we have 3(0) − 3𝑦 = 9 ⇒ −3𝑦 = 9. Divide both sides by −3 to get 𝑦 =
−3. Thus, the y-intercept is (0, -3). The line through these two points is plotted above. Note that this line is
dotted, since the original inequality used a > sign rather than a ≥ sign.
Now we must decide which side of the line to shade. We can do this by picking any point not on the line and
seeing whether the inequality holds for this point. Here, we can pick (0,0) for convenience. 3(0) − 3(0) − 9 <
0 ⇒ −9 < 0 is true, so the inequality holds for the point (0, 0) and we shade the side of the graph that
contains (0, 0).
21. Write the equation of a horizontal line passing through (-8,-5).
Solution: A horizontal line has the form 𝑦 = 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟. Since this horizontal line goes through a point with ycoordinate −5, it has equation 𝑦 = −5.
22.
(−3𝑥 4 𝑦 −2 )−3 (𝑥 3 6𝑦 4 )0
Solution: (−3𝑥 4 𝑦 −2 )−3 (𝑥 3 6𝑦 4 )0 = (−3𝑥 4 𝑦 −2 )−3 (1) = (−3𝑥 4 𝑦 −2 )−3, since any expression raised to the 0th
power is equal to 1.
−3𝑥 4 −3
) ,
𝑦2
(−3𝑥 4 𝑦 −2 )−3 = (
since expressions in the numerator raised to negative powers can
−3𝑥 4 −3
)
𝑦2
be moved to the denominator, switching the sign of the exponent to positive. (
since a fraction to the -1st power is the reciprocal of the original. Now, distribute the third
−
23.
−3𝑥 4 −1 3
𝑦2 3
)
]
=(
) ,
2
𝑦
−3𝑥 4
2
𝑦
power to get (−3𝑥4 )3 =
= [(
𝑦6
27𝑥 12
(8𝑦 − 2 + 3𝑦 3 ) − (12𝑦 + 10 − 9𝑦 3 )
Solution: (8𝑦 − 2 + 3𝑦 3 ) − (12𝑦 + 10 − 9𝑦 3 ) = 8𝑦 − 2 + 3𝑦 3 − 12𝑦 − 10 + 9𝑦 3 = 12𝑦 3 − 4𝑦 − 12
24. (6𝑥 − 5𝑦)(𝑦 + 4𝑥)
Solution: (6𝑥 − 5𝑦)(𝑦 + 4𝑥) = 6𝑥𝑦 − 5𝑦 2 + 24𝑥 2 − 20𝑥𝑦 = 24𝑥 2 − 14𝑥𝑦 − 5𝑦 2
25. (3𝑥 − 8𝑦)2
Solution: (3𝑥 − 8𝑦)2 = (3𝑥 − 8𝑦)(3𝑥 − 8𝑦) = 9𝑥 2 − 24𝑥𝑦 − 24𝑥𝑦 + 64𝑦 2 = 9𝑥 2 − 48𝑥𝑦 + 64𝑦 2
1
1
26. (4 𝑦 2 − 2𝑧) (4 𝑦 2 + 2𝑧)
1
2
1
Solution: By the difference of two squares formula, the product is (4 𝑦 2 ) − (2𝑧)2 = 16 𝑦 4 − 4𝑧 2
27. 12𝑧 3 (6𝑧 4 − 3𝑧 − 4)
Solution: Use the distributive law:
12𝑧 3 (6𝑧 4 − 3𝑧 − 4) = (12𝑧 3 )(6𝑧 4 ) − (12𝑧 3 )(3𝑧) − (12𝑧 3 )(4) = 72𝑧 7 − 36𝑧 4 − 48𝑧 3
5
28. Determine the GCF of the following terms. 24𝑥 4 𝑦 3 , −54𝑥 3 𝑦 3 , 15𝑥 3 𝑦 5
Solution: Since 24𝑥 4 𝑦 3 = 23 ∙ 3𝑥 4 𝑦 3 , −54𝑥 3 𝑦 3 = −2 ∙ 33 𝑥 3 𝑦 3 , 15𝑥 3 𝑦 5 = 3 ∙ 5𝑥 3 𝑦 5 , the GCF is 3𝑥 3 𝑦 3 .
29. 𝑥 3 + 6𝑥 2 − 3𝑥 − 18
Solution: Since this has four terms, try factoring by grouping. 𝑥 3 + 6𝑥 2 − 3𝑥 − 18 = (𝑥 3 + 6𝑥 2 ) + (−3𝑥 − 18) =
𝑥 2 (𝑥 + 6) + (−3)(𝑥 + 6). Factoring out an (𝑥 + 6), we get (𝑥 + 6)(𝑥 2 − 3)
30. 𝑥 2 + 7𝑥 − 30
31. 3𝑦 3 + 33𝑦 2 − 78𝑦
32. 9𝑧 2 + 3𝑧 − 20
33. 25𝑥 2 − 49𝑦 2
34. 16𝑥 2 − 8𝑥 + 1
35. Solve: 8𝑥 2 + 6𝑥 = 9
Define a variable, use it in an equation and solve.
36. The product of two consecutive positive integers is 156. Find the integers.
Solution: Let 𝑥 be the smaller of the two integers. Since they are consecutive, the larger can be represented as 𝑥 +
1. Then the sentence can be translated to 𝑥(𝑥 + 1) = 156. Converting this to our usual form for solving quadratics,
we get 𝑥(𝑥 + 1) = 156 ⇒ 𝑥 2 + 𝑥 = 156 ⇒ 𝑥 2 + 𝑥 − 156 = 0. To factor this, we need to find some 𝑎, 𝑏 such
that (𝑥 + 𝑎)(𝑥 + 𝑏) = 𝑥 2 + 𝑥 − 156. Note: The x’s in the factors do not have coefficients since the leading
coefficient is 1. Through guess and check, we get (𝑥 + 13)(𝑥 − 12) = 𝑥 2 + 𝑥 − 156 = 0.
Thus 𝑥 + 13 = 0 ⇒ 𝑥 = −13 or 𝑥 − 12 = 0 ⇒ 𝑥 = 12. Since it is stated in the problem that the integers are
positive, we can conclude that 𝑥 = 12. So the two consecutive integers are 12 and 13. To check, you can verify that
the product of 12 and 13 is indeed 156.
37. The length of a rectangle is 5 feet more than the width. The area is 176 square feet. Find the dimensions.
Solution: Let 𝑥 be the width of the rectangle. Since the length is 5 feet more than the width, the length can be
represented by 𝑥 + 5. Since the area of a rectangle is the product of the length and width, 𝑥(𝑥 + 5) = 176 ⇒
𝑥 2 + 5𝑥 = 176 ⇒ 𝑥 2 + 5𝑥 − 176 = 0. This can be factored through guess and check to get
(𝑥 + 16)(𝑥 − 11) = 0. So either 𝑥 + 16 = 0 ⇒ 𝑥 = −16 or 𝑥 − 11 = 0 ⇒ 𝑥 = 11. Since the dimensions of a
rectangle must be positive, we can conclude that 𝑥 = 11. So the width is 11 feet. Since the length is 5 feet more
than the width, the length is 11 + 5 = 16 feet.
38. Find the domain of
𝑥−1
𝑥 2 +4𝑥
Solution: This expression is defined everywhere except the x-values for which the denominator is 0. To find these
values, solve 𝑥 2 + 4𝑥 = 0. Factoring out an 𝑥, we get 𝑥(𝑥 + 4) = 0. So the denominator is 0 when 𝑥 = 0 and when
𝑥 + 4 = 0 ⇒ 𝑥 = −4. Thus the domain is 𝑥 ≠ 0, 𝑥 ≠ −4.
6
10𝑥 2 −21𝑥−10
39. Simplify
2𝑥 2 −𝑥−10
Solution:
10𝑥 2 −21𝑥−10
2𝑥 2 −𝑥−10
5
Note that 𝑥 ≠ 2 , −2.
5
restriction 𝑥 ≠ 2.
40.
7
−
4𝑦 2
=
(5𝑥+2)(2𝑥−5)
(2𝑥−5)(𝑥+2)
=
5𝑥+2
𝑥+2
Since it is clear from the final answer that 𝑥 ≠ −2, we need only explicitly state the domain
5
𝑦
Solution: Before adding, we must manipulate the expression so that both fractions have the same denominator. In
this case, we can see that a possible common denominator is 4𝑦 2 , since this expression is divisible by both of the
denominators. This is actually the smallest possible common denominator, which is convenient but not necessary.
The first fraction already has this as its denominator, so we only have to convert the second fraction. We can do this
by multiplying both the numerator and the denominator by 4𝑦.
7
−
4𝑦 2
41.
42.
5
𝑦
=
7
4𝑦 2
−
20𝑦
4𝑦2
=
5
𝑦
∗
4𝑦
4𝑦
=
20𝑦
4𝑦2
. So we have
7−20𝑦
4𝑦 2
𝑥−5 𝑥 2 +𝑥−30
𝑥+6
∙
𝑥
1−3𝑥
𝑥 2 −25
÷
𝑥 2 −1
3𝑥 2 +2𝑥−1
Solve. Check for extraneous solutions.
43.
44.
5
8𝑥
2
−11
𝑥
24
− =
1
𝑥 2 −2𝑥−15
+
−6
𝑥−5
=
4
𝑥+3
Define a variable, use it in an equation, show all steps in solving the equation, and label your answer.
45. Brett takes 4 hours to paint a room and Devon takes 6 hours. How long will it take them working together?
46. The difference of an integer and twice its reciprocal is
47
.
7
Find the number.
7
Solve each system of linear equations. If a system has no solution or infinitely many solutions, state so.
8𝑥 + 2𝑦 = 3
−4𝑥 − 𝑦 = 1
47.
𝑥 = 3𝑦 + 10
2𝑥 − 3𝑦 = 14
48.
2𝑥 − 7𝑦 = 30
5𝑥 − 𝑦 = 9
49.
Define your variables, write a system of linear equations, show all steps in solving, and label your answer.
50. The difference of two numbers is 8. The smaller number is one-third the larger. What are the numbers?
51. A deli sells 80 club and BLT sandwiches, for a total of $ 448. If a club sandwich costs $ 5.00 and a BLT sandwich
costs $ 6.00, how many of each type of sandwich did they sell?
52. A total of $18,000 is invested in two bonds paying 2% and 1.5% simple interest. The annual interest is $260. How
much is invested in each bond?
Simplify. Use absolute value signs if necessary. DO NOT use a calculator to approximate values.
53. √40
54. √50𝑥 4 𝑦 7
3
55. √−54𝑥 5 𝑦 3
Rationalize the denominator.
5
56. √6
5
Solution: √6 =
√5
√6
=
√5 √6
∙
√6 √6
=
√30
5
8
55.
√3
√12−2
56. −3√48 + √3 − 8√6 − 9√75
Solution: −3√48 + √3 − 8√6 − 9√75 = −3√16 ∙ 3 + √3 − 8√6 − 9√25 ∙ 3 = −3 ∙ 4√3 + √3 − 8√6 − 9 ∙ 5√3
= −12√3 + √3 − 8√6 − 45√3 = −56√3 − 8√6
57. 4√10(√5 − 2√10)
58. (√2 − 3)(5√2 + 7)
59. (3𝑦 − √11)(3𝑦 + √11)
2
60. (𝑥 − √5)
61. Find the measure of the diagonal of a rectangle whose length is 10 cm and whose width is 14 cm. Find the exact
measure and then use your calculator to round your answer to 3 decimal places.
Solve the following quadratic equations. Write any complex solutions in standard form.
62. 6𝑦 2 − 24𝑦 = 0
Solution: Since there is no constant term, this quadratic equation can easily be solved by factoring. Factoring out the
GCF gives: 6𝑦(𝑦 − 4) = 0. By the zero product property, 6𝑦 = 0 ⇒ 𝑦 = 0 or 𝑦 − 4 = 0 ⇒ 𝑦 = 4. Thus the
solutions are 𝑦 = 0, 4.
63. 𝑥 2 − 6𝑥 = 15
64. 2𝑥 2 − 𝑥 = −8
Define a variable, use it in an equation, show all steps in solving the equation, and label your answer.
65. A picture is 7 inches shorter than it is long and has an area of 78 square inches. Find the dimensions.
Solution: Let 𝑥 be the length of the picture. The width is then given by 𝑥 − 7. Since the area is 78 square inches, we
have that 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ = 78 or 𝑥(𝑥 − 7) = 78 ⇒ 𝑥 2 − 7𝑥 = 78 ⇒ 𝑥 2 − 7𝑥 − 78 = 0 ⇒ (𝑥 − 13)(𝑥 + 6) = 0
By the zero product property, either 𝑥 = 13 or 𝑥 = −6. Since a length must be positive, we can conclude that the
length of the picture is 13 inches. Thus the width is 13 − 7 = 6 inches. Thus the dimensions are 13 inches by 6
inches.
Perform the operations and write the result in standard form.
66. (8 + 7𝑖)— (9 − 2𝑖 )
9
Solution: (8 + 7𝑖)— (9 − 2𝑖 ) = 8 + 7𝑖 − 9 + 2𝑖 = −1 + 9𝑖
67. √−36(5 − √−81)
Solution: √−36(5 − √−81) = 6𝑖(5 − 9𝑖) = 30𝑖 − 54𝑖 2 = 30𝑖 − 54(−1) = 54 + 30𝑖
68. (4 − 49𝑖)(4 + 49𝑖)
69. (3𝑖 + 8)2
Solution: (3𝑖 + 8)2 = (3𝑖 + 8)(3𝑖 + 8) = 9𝑖 2 + 24𝑖 + 24𝑖 + 64 = 9(−1) + 48𝑖 + 64 = 55 + 48𝑖
For the equation 𝑦 = 𝑥 2 − 4𝑥 + 3 find:
70. Vertex
71. X-intercepts
72. Y-intercepts
73. Sketch the graph.
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