Review Solutions
1. Solve. Simplify your answer as much as possible.
−21 = − 37 w
78 − u = 168
3y − 8 = −20
− 90 = u
49 = w
−5u − 18 = −2(u − 6)
u = −10
w−4 w+1
−
=1
5
3
w = −16
2. Solve for L.
−2 =
y = −4
w
w
+4=
3
2
24 = w
−
3x − 23
4
5=x
3
2
9
=− u−
2
7
5
21
− 20 = u
3(w − 2) − 5w = −2(w + 3)
Solve for B.
A = 4B + C
Solve for x.
y = (x − 8)m
y = −11
9=
9y + 5 y − 6
+
8
2
y=7
5(2 − v) + 7v = 2(v + 1)
all Real numbers
4LW = V
23 + 18y = −21 + 14y
no solution
V
4W
A−C
=B
4
y
+8=x
m
L=
3. A house was increased in value by 34% since it was purchased. The current value is $335,000. What was the
value when it was purchased?
1.34x = 335, 000
x = 250, 000
so it was original purchased for $250,000.
4. A TV has a listed price of $503.98 before tax. If sales tax is 7.5%, find the total cost of the TV with sales
tax included.
1.075(503.98) = 541.79 The price with tax is $541.79.
5. The perimeter of a rectangle is 132 feet. If the length is represented by 5y and the width by 4y + 3, find the
length and width in feet.
L = 5y
W = 4y + 3
P = 2W + 2L and P = 132 so 132 = 2(5y) + 2(4y + 3), so by solving this, y = 7.
This means L = 5 · 7 = 35 and W = 4 · 7 + 3 = 31.
6. Solve the inequality. Graph your solution and write it in interval notation.
−4u + 37 ≤ 13
8w − 40 > −3(4 − 5w)
− 12 ≤ 4x + 4 < 16
Remember when solving inequalities, if you multiply or divide by a negative number this will change the
sign of the inequality.
u≥6
[6, ∞)
−4>w
(−∞, −4)
−4≤x<3
[−4, 3)
7. Solve the following absolute value equations or inequalities. For the inequalities, graph your solutions and
write your solutions in interval notation.
Remember when solving equations with absolute values, after you isolate the absolute value, you need to
keep in mind that you can have a positive and negative version of the other side. For example, on the first
problem, 6v − 3 = 9 but also 6v − 3 = −9.
|6v − 3| = 9
v = 2, −1
|4u + 2| − 35 = −5
u = 7, −8
|5v − 9| = |5v + 4|
v=
1
2
When solving the inequalities, you can treat them as if they are equalities (= in place of < or >), and then
place your solutions on a number line and test which parts of the number line work in the original inequality.
For example, on the second problem below, when you solve with an = sign, you get x = −4 and x = 8. When
you place these on a number line, it breaks the number line into 3 parts, (−∞, −4), (−4, 8), and (8, ∞). You
can pick one point from each of these intervals and test them into the original inequality. When you do this,
you see that any numbers in the intervals (−∞, −4) or (8, ∞) work in the original inequality, but the other
interval does not work.
|4x + 4| ≤ 8
|u − 2| > 6
(−∞, −4) ∪ (8, ∞)
[−3, 1]
8. Find the x-intercept and y-intercept for the following, and then use them to graph the functions.
2x + 4y = −8
y = − 14 x + 2
x-intercept: (−4, 0) y-intercept: (0, −2)
x-intercept: (8, 0) y-intercept: (0, 2)
9. Find the slope and y-intercept, then graph.
−3x − 5y = −15
slope: m =
− 35
y-intercept: b = 3
− 3x + 4y = −20
slope: m =
3
4
y-intercept: b = −5
10. Find an equation for the line described. You can leave your answer in point-slope form or slope-intercept
form.
Line passing through the points (−9, −6) and (−4, 5). y − 5 =
11
5 (x + 4)
or y + 6 =
11
5 (x + 9)
or y =
11
69
5 x+ 5
Line passing through (−8, 2) with a slope of − 54 . y − 2 = − 54 (x + 8)
Line passing through (9, 6) and parallel to the line y = 32 x. y − 6 = 32 (x − 9)
Line passing through the origin and perpendicular to the line y = 32 . Since this is y = a constant, this is
a horizontal line. So the line perpendicular to it must be a vertical line, and since it must go through the
origin, it is x = 0.
11. Owners of a recreation area are filling a small pond with water at a rate of 35 liters per minute. There are
700 liters in the pond when they begin.
Let W represent the amount of water in the pond (in liters), and let T represent the number of minutes the
water has been added.
Write an equation relating W to T , and the graph your equation.
W = 35T + 700
12. The entire graph of the function h is shown in the figure below. Write the domain and range of h using
interval notation.
Domain: [−2, 3) Range: [−4, 4)
13. The graph of a function f is shown below.
Find f (2) and find one value of x for which f (x) = −3.
f (2) = −2 notice when x = 2, y = −2.
f (x) = −3 when x = −1.5 or x = 1.5 notice when x = 1.5, −1.5, y = −3.
14. Find the domain of the function, and write your solution in interval notation.
√
√
√
g(x) = x − 4
f (x) = −x + 9
h(x) = 2x − 3
x − 4 ≥ 0 so x ≥ 4 [4, ∞)
−x + 9 ≥ 0 9 ≥ x (−∞, 9]
2x − 3 ≥ 0 x ≥
3
2
[ 23 , ∞)
15. Solve the following system of equations.
y = 3x − 4
4x + 3y = 27
6x + 9y = −3
7x − 2y = −9
6x + 5y = 9
4x − 5y = −9
You can use direct substitution on the first set. You can subtract the second set. For the third set, one
method would be to multiply the first equation by 5 and the second equation by -2 and then add the two
equations.
x = 4 y = −3 (4, −3)
x = 3 y = 5 (3, 5)
x = −1 y = 1 (−1, 1)
16. A jet travels 1464 miles against the wind in 2 hours, and it travels 1704 miles with the wind in the same
amount of time. What is the rate of the jet in still air, and what is the rate of the wind?
Remember, distance=rate · time, so D = R · T . Since you are asked to find the rate, you can solve for R,
giving R = D
T.
If you let x represent the jet’s rate in still air, and y represent the wind’s rate, then when the jet is flying
against the wind, the jet’s rate is reduced by the wind’s rate. so against the wind: x − y = 1464
2 = 732 so
x − y = 732.
When the jet is flying with the wind, the jet’s rate is increased by the wind’s rate, so with the wind:
x + y = 1704
2 = 852 so x + y = 852.
Now you can add the two equations and solve for x, and then sub back in to find y. This gives the jet’s rate
to be 792 miles per hour, and the wind is 60 miles per hour.
17. Hong bought a desktop computer and a laptop. Before financing charges, the laptop cost $400 less than the
desktop. He paid for the computers using two different financing plans. For the desktop, the interest rate
was 7.5% per year, and for the laptop, it was 8% per year. The total finance charge for one year was $371.
How much did each computer cost before finance charges?
If L is the price of the laptop, and D is the price of the desktop, the fact that the laptop is 400 less than
the desktop gives L = D − 400. Then using the interest rates and the total finance charge (interest charge),
we get 0.075D + 0.08L = 371. Now you can sub the first equation into the second and solve. The desktop
is $2600 and the laptop is $2200.
18. Simplify as much as possible. Leave your answers with only positive exponents.
!3
−
4a
6u4 v 2
3 z 4 )2 (2x2 y 3 z)
2y 5 v 5 · 4v 4 · 6y
(−7ab3 )2
(−x
b3
18uv 2
48v 9 y 6
4m4
3m7 n2
49a2 b6
−
64a3
b9
2x8 y 3 z 9
!2
16
9m6 n4
4v −4
4
v4
−
27w12
x6
1
81
3u3
!3
2x−3 u
(x2 z −1 )
z −2
36x−4 y 3 z −2
6y −5 z 6
(−3w4 x−2 )3
(−9)−2
6y 8
x4 z 8
8u3 z 5
x7
19. Factor.
2w3 + 5w2 + 14w + 35
(2w+5)(w2 +7)
ux − 7x − 3u + 21
y 2 − 10y + 16
(u−7)(x−3)
(y−8)(y−2)
4y 2 − 28y + 48
3y 2 − 4y − 7
4(y−3)(y−4)
(3y−7)(y+1)
3y 2 − 4y − 20
(3y−10)(y+2)
y 2 − 14y + 49
(y−7)(y−7)
27w3 − 48w
3w(3w+4)(3w−4)
w2 − 36
(w+6)(w−6)
32u2 − 2u2 v 4
2u2 (4+v 2 )(2+v)(2−v)
20. Solve.
3y 2 − 18y = 0
u2 − 10u + 21 = 0
3y(y − 6) = 0 y = 0, 6
5w2 = −17w − 6
(5w + 2)(w + 3) = 0 w = − 25 , −3
(u − 7)(u − 3) = 0 u = 7, 3
21. Find all values of x that are NOT in the domain of the following functions.
3x
x2 + 2x − 63
g(x) =
2x − 10
x2 − 49
To answer these, we can not let the denominators be zero, so any x value that makes the denominator zero
is NOT in the domain.
f (x) =
x2 − 49 = 0 x = ±7 are not in the domain.
2x − 10 = 0 x = 5 is not in the domain.
22. Simplify.
4u2 − 100
u2 − 8u + 15
x−8
x2 − 64
7(2w + 5)(w + 6)
21(w + 4)(2w + 5)
w+6
3(w + 4)
=
x−8
1
=
(x + 8)(x − 8) x + 8
=
4(u + 5)(u − 5) 4(u + 5)
=
(u − 5)(u − 3)
u−3
23. Perform the operation and simplify.
4x − 20 9x − 8
·
45x − 40 2x − 10
2y 9ay
·
3a 10y 5
3
5y 3
4(x−5)
9x−8
5(9x−8) · 2(x−5)
=
=
2
5
c2 − 9c
3c + 8
+
c2 − c − 12 c2 − c − 12
=
c2 −6c+8
c2 −c−12
(c−4)(c−2)
(c−4)(c+3)
=
=
·
9t2 u
5rs2
=
=
4(x+2)
x−1
(x−3)(x+2) · x−2
5
9
− 2
6y 8y
20y
27
− 24y
2
24y 2
=
=
20y−27
24y 2
=
=
4(x−1)
(x−3)(x−2)
=
(x−2)(x−1)
x+2
(x+3)(x+2) · 4(x−2)
=
4(x−1)+2(x+1)
(3x−1)(x−1)(x+1)
3
x + 6 · (x + 6)
(x + 6)
9
+x
x+6
=
=
x+6−3
9+x2 +6x
x+3
(x+3)(x+3)
=
x−1
4(x+3)
2
(x+1)(x−1)
1−
35y−2
15y−2
=
4
2
+
3x2 + 2x − 1 3x2 − 4x + 1
23−x
(x−5)(x+4)
7−
9s2
rt3 u2
=
2
3
−
x−5 x+4
2
5y 5y
·
2 5y
3−
5y
15s4
3t5 u3
5rs2
9t2 u
15s4
3t5 u3
c−2
c+3
x2 − 3x + 2 4x − 8
÷
x2 + 5x + 6
x+2
x−1
4x + 8
·
x2 − x − 6 x − 2
=
u−1
1
x+3
=
1
1
−v 2
u
·
u
u
1
− v2
=
u
1−uv 2
24. Evaluate
1
1
256 4
=
√
4
1
27 3
256 = 4
=
√
3
27 = 3
1
(−8) 3
81− 4
√
3
1
1
= √
=
4
81 3
=
−8 = −2
25. Simplify.
1
5
7
w ·w
20
21
13
w 28 + 28 = w1 28 = w
√
√
27u14
u2
3
4
√
28
u
w13
√
75 − 3 27
√
√
√
= 5 3 − 9 3 = −4 3
7
12
6
7
5
u 14 − 14 = u− 14 =
1
√
14
u5
√
√
4z 32z + 18z 3
√
√
√
= 16z 2z + 3z 2z = 19z 2z
√
√
3u7 3
√
√
3x6 6x
5z ·
√
3
=
√
3
3 · 4u2 ·
40t8 w3
√
3
2t2 w 5t2
√
2t2 y 4 2t
√
7z
√
= z 35
√
3
p
8t5 y 8
54x13
√
3
32 u5
√
33 · 4u7 = 3u2 3 4u
√ √
√
√
( x − 2)( x + 2)
√
√
x+ 2x− 2x+2 = x−2
√
(x + 2 2)2
√
√
√
x2 +2x 2+2x 2+8 = x2 +4x 2+8
26. Solve. Remember to check your solutions.
√
√
√
3y + 18 + 2 = 5
5x + 10 = 7x − 12
√
√
4
1
2
3
y·
p
3
y2
8
11
y 4 + 3 = y 12 + 12 = y 12 =
11y − 30 = y
u−5=
√
p
12
y 11
49 − 8u
Isolate the radical, and then square both sides of the equation. Check your solutions. Most of these work,
but the last problem has two possible solutions u = 6, −4, but when you check them, only u = 6 is the
correct solution.
y = −3
11 = x
y = 5, 6
u=6
27. Perform the operation and simplify. Write your solution in a + bi form.
(6 − 2i) + (4 + 3i)
(3 − 7i) − (5 + 4i)
4 − 2i 2 + 5i
·
2 − 5i 2 + 5i
(−3 + 6i)(−4 + 3i)
For the last two problems, you will use the fact that i2 = −1 to simplify them.
10 + i
12 − 9i − 24i + 18i2 = −6 − 33i
− 2 − 11i
8+20i−4i−10i2
4−25i2
=
18
29
+ 16
29 i
28. Solve. You may need to use the quadratic formula.
(v − 7)2 − 32 = 0
√
v =7±4 2
(w + 9)2 − 45 = 0
√
w = −9 ± 3 5
2x2 + 5x − 1
x=
√
−5± 33
4
The third problem had a typo. It should’ve been set equal to zero.
4x2 − 9x + 3 = 0
3x2 + 5x = 3
x=
x=
√
9± 33
8
2x2 − 3x + 6 = 0
√
−5± 61
6
x=
29. A model rocket is launched with an initial upward velocity of 235
seconds is given by h = 235 − 16t2 .
ft
sec .
√
3±i 39
4
The rocket’s height h (in feet) after t
Find all values of t for which the rocket’s height is 151 ft.
This problem also had a typo. The equation should’ve been h = 235t − 16t2 . However, since the original
does not say that, I will give the solution according to the original.
151 = 235 − 16t2
− 84 = −16t2
21
4
= t2
√
21
2
=t
30. Graph the following functions. Make sure to label the vertex.
g(x) = −2x2
h(x) = 3x2 − 1
y = (x − 1)2 − 3
31. State the vertex and x-intercepts of the following functions, then use them to graph the function.
y = x2 − 4x − 21
y = x2 − 8x + 12
Use the fact that the x-value of the vertex is −b
2a , and then plug this back in to get the y-value of the vertex.
To find the x-intercepts, you can set y = 0, and then factor and solve for x.
vertex: (2, −25) x-intercepts: (7, 0), (−3, 0)
vertex: (4, −4) x-intercepts: (6, 0), (2, 0)
32. Given f (x) = −2x2 + 16x − 34, answer the following.
Does the function have a minimum or a maximum value? Maximum, because the negative coefficient of x2
means it opens downward, making the vertex a maximum.
At what x value does the min/max occur? x =
−16
2(−2)
= 4 so x = 4.
What is the min/max value? y = f (4) = −2(42 ) + 16(4) − 34 so y = −2
33. A supply company manufactures copy machines. The unit cost C (cost in dollars to make each copy machine)
depends on the number of machines made. If x machines are made, then the unit cost is given by C(x) =
0.5x2 − 170x + 25, 850. What is the minimum unit cost?
Since this is a quadratic equation, with a leading coefficient that is positive, it opens upward, making its
vertex a minimum. To find the minimum unit cost, we need to find the y-value of the vertex.
x=
170
2(0.5)
= 170 C(170) = 0.5(1702 ) − 170(170) + 25, 850 = 11, 400 so the minimum is $11, 400.
34. s(x) = 3x + 6
t(x) = 4x
u(x) = x2 + 7
w(x) =
√
x+8
Given the functions defined above, find the following expressions.
(s · t)(x)
(s + t)(x)
= 3x + 6 + 4x = 7x + 6
w(u(x))
= w(x2 + 7) =
√
x2 + 7 + 8 =
= (3x + 6)(4x) = 12x2 + 24x
√
x2 + 15
(s − t)(4)
(s − t)(x) = −x + 6 so (s − t)(4) = −4 + 6 = 2
w(u(1))
√
√
= 12 + 15 = 16 = 4
u(w(1))
√
u( 1 + 8) = u(3) = 32 + 7 = 16
35. For each pair of functions below, find f (g(x)) and g(f (x)). Then determine whether f and g are inverses of
each other.
6
x+7
f (x) =
f (x) = 2x + 3
f (x) =
x
5
6
g(x) =
g(x) = 2x − 3
g(x) = 5x − 7
x
To show that two functions are inverses of one another, you need to show that f (g(x)) = x as well as
g(f (x)) = x.
f (g(x)) =
g(f (x)) =
6
6
x
6
6
x
= 61 · x6 = x
= 61 · x6 = x
f (g(x)) = f (2x−3) = 2(2x−3)+3 = 4x−3 6= x
notice this is not an inverse from f (g(x)) 6= x.
f (g(x)) = f (5x−7) =
h(x) = 5x3 + 7
=x
5 x+7
g(f (x)) = g( x+7
5 ) = 1 ( 5 )−7 = x
36. h is a one-to-one function. Find h−1 (x).
h(x)4x + 3
5x−7+7
5
h(x) =
√
3
2x + 5
There was a typo on the first problem. It should read h(x) = 4x + 3. To find an inverse function, you can
(1) set f (x) = y, then (2) switch your x and y variables, and then (3) solve for y. This is now your f −1 (x).
I’m am going to start with step 2 being done here.
√
x = 4y + 3 solve for y, &
x = 5y 3 + 7 solve for y, &
x = 3 2y + 5 solve for y, &
q
x−3
x3 −5
3 x−7
−1 (x)
−1
=
h
= h−1 (x)
4
5 = h (x)
2
37. f is a one-to-one function, f (x) =
√
x + 5 + 4.
Find the domain and range of f (x). Then find f −1 (x) and its domain.
Domain of f : x ≥ −5 or you can write [−5, ∞)
Range of f : y ≥ 4 or [4, ∞) notice the lowest number the square root can be is zero and if we then add 4,
the lowest y value possible for this function is 4.
√
To find the inverse: x = y + 5 + 4 solve this for y, (x − 4)2 − 5 = y so f −1 (x) = (x − 4)2 − 5
Remember that inverse functions swap domains and ranges. So the domain of f −1 is simply the Range of f ,
so [4, ∞) .