MAT 062
PRACTICE FINAL EXAM SOLUTIONS
1. Simplify the expression 3x − 2y − 4 ( 2x − 6y + 3z )
!
3x − 2y − 8x + 24y − 12z
−5x + 22y − 12z
2. Evaluate 3x 2 − 2 ( 3x − 2y ) for x = −1 and y = 2
3( −1) − 2 ⎡⎣ 3( −1) − 2 ( 2 ) ⎤⎦
2
3(1) − 2 ( −3 − 4 )
!
3 − 2 ( −7 )
3 + 14
17
3. Translate “3 more than the product of a number and 5” into an algebraic expression.
!
5n + 3
4. Solve the equation 7x + 3( 4x − 5 ) = 8x − 26 for x.
7x + 12x − 15 = 8x − 26
19x − 15 = 8x − 26
!
11x − 15 = −26
11x = −11
x = −1
5. Last year 1,500 cars were sold at a particular dealership. This was a decrease of
6% from the previous year. How many cars were sold that year? Round to the
nearest whole number if necessary.
1500 = x − 0.06x
!
1500 = 0.94x
x = 1596
!
1596 cars sold in the previous year
6. Solve the formula F = Amn − Bn for m
F = Amn − Bn
!
F + Bn = Amn
F + Bn
=m
An
7. A certain cell phone company charges $25 per month plus $0.05 per minute of use.
If Sally has a budget of $32, what is the maximum number of minutes she can use
her phone and stay within budget? In solving this problem, the first step is to define
a variable. What is the most appropriate definition for this problem?
a. Let x represent the cost of the phone for one month
b. Let x represent the number of days to use the phone
c. Let x represent the number of minutes used
d. Let x represent the cost of using one minute
e. Let x represent Sallyʼs budget
!
c. Let x represent the number of minutes used
8. Solve the problem in #7. Round your answer to the nearest minute.
Cost = 25 + 0.05x
32 = 25 + 0.05x
!
7 = 0.05x
x = 140
!
She can use up to 140 minutes and stay within budget
3
9. Solve the equation x − 2 = 4 for x.
4
3
x−2=4
4
3x − 8 = 16
!
3x = 24
x=8
10. Solve 16 − 3x ≥ 7 . Use set builder notation to write your solution.
16 − 3x ≥ 7
−3x ≥ −9
!
x ≤ 3 (dividing by negative changes the inequality sign)
{ x x ≤ 3}
11. Find the intercepts of the equation 3x − 9y = 18 .
x-intercept (y = 0): 3x − 9(0) = 18
3x = 18
x = 6 (6, 0)
y-intercept (x = 0): 3(0) − 9y = 18
−9y = 18
y = −2 (0, −2)
12. David rented a car for $150. During his 2 day rental, he drove 300 miles. Find his
average rate of change in miles per dollar.
300miles
!
= 2 miles dollar
$150
13. Find the slope of the line containing the points (2, 4) and (-2, 6).
6−4
m=
−2 − 2
2
!
m=
−4
1
m=−
2
14. Write the equation of the line with slope m = 3 and passing through the point (1, -5).
y − (−5) = 3(x − 1)
!
y + 5 = 3x − 3
y = 3x − 8
15. Write a linear equation to represent the data in the table.
x
y
0
2
3
0
6
-2
y-intercept is 2 (from table) and m =
2−0
2
=−
0−3
3
2
so equation is y = − x + 2
3
16. Find the equation of the vertical line which is to the right 2 spaces from the y-axis.
!
vertical lines have equations of the form x = a, so this one is x = 2
17. Graph the equation y = 3x + 4 .
!
(description): y-intercept of 4, slope of 3 so fairly steep going up left to right
!
another point is (7, 1)
18. Simplify 2x 3 + 3x 2 − 1 − 5x 3 − 6x 2 − 1
(
!
) (
2x 3 + 3x 2 − 1 − 5x 3 + 6x 2 + 1
−3x 3 + 9x 2
(
19. Multiply ( x + 2 ) 3x 2 − 5x − 8
!
)
)
3x 3 − 5x 2 − 8x + 6x 2 − 10x − 16
3x 3 + x 2 − 18x − 16
20. Simplify ( 2x − 6 )
!
2
4x 2 − 24x + 36
(
)
21. Perform the indicated operation 3x 3 + x 2 − 18x − 16 ÷ ( x + 2 )
!
!
long division, I canʼt figure out how to include
answer: 3x 2 − 5x − 8
⎛ y3 ⎞
22. Simplify ⎜ ⎟
⎝ 2⎠
!
⎛ y3 ⎞
⎜⎝ 2 ⎟⎠
−2
−2
2
⎛ 2⎞
4
= ⎜ 3⎟ = 6
⎝y ⎠
y
23. Multiply and write the answer in scientific notation.
( 3.2 × 10 )( 6.5 × 10 ) = ( 3.2 )( 6.5 ) × 10
2
!
−5
( 3.2 × 10 )( 6.5 × 10 )
2
−5
2−5
= 20.8 × 10 −3
= 2.08 × 10 −2
24. Find the greatest common factor in the expression 48x 7 y 8 − 12x 5 y 3 + 36x 3 y 5
!
GCF: 12x 3 y 3
25. Find b so that x 2 + bx − 10 can be factored.
!
Considering factors of 10 and their differences, b could either be 9 (for 1 and 10)
!
or 3 (for 2 and 5)
26. Factor 2x 2 − 18x + 40 completely.
2(x 2 − 9x + 20)
!
2(x − 5)(x − 4)
27. Solve the equation 3x 2 + 7x = 26
3x 2 + 7x − 26 = 0
( 3x + 13) (x − 2) = 0
!
3x + 13 = 0 or x − 2 = 0
3x = −13
x=2
13
x=−
3
28. Write an appropriate equation to solve the following problem (DO NOT SOLVE):
The height of a triangle is 6 inches more than twice its width. If the area of the
triangle is 20 square inches, find the width and height.
1
A = bh
2
!
1
20 = w ( 2w + 6 )
2
5
7
29. Perform the indicated operation.
+
8y 12y 2
LCD: 24y 2
5 ⎛ 3y ⎞
7 ⎛ 2⎞
+
⎜ ⎟
⎜
⎟
8y ⎝ 3y ⎠ 12y 2 ⎝ 2 ⎠
!
15y
14
+
2
24y
24y 2
15y + 14
24y 2
x 2 − 7x − 18
30. Simplify
if possible.
x2 − 4
( x − 9 )( x + 2 )
( x + 2 )( x − 2 )
!
x−9
x−2
2x − 10 14x 2
31. Multiply and simplify.
⋅
18x 3x − 15
!
2 ( x − 5 ) 14x 2
⋅
18x
3( x − 5 )
14x
27
32. Perform the indicated operation and simplify.
!
x 2 + 2x − 8
x+4
( x + 4 )( x − 2 )
x+4
x−2
x 2 + 2x
8
−
x+4
x+4