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Math 101 10–9–
Review Sheet for Test 2
These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence of a topic does not imply that it won’t appear on the test.
- (a) Show that it is not valid to cancel x’s in
x + 4 x to get
= 5 by giving a specific value of x for
which
x + 4 x is not equal to 5.
(b) Show that
a b c
is not always the same as
a b c
by giving specific values of a, b, and c for which
a b c
is not
equal to
a b c
- Simplify, cancelling any common factors:
(a) x^2 − x − 2 x^2 + 3x + 2
x^2 − 5 x + 6 x^2 − 4 x + 4
(b)
x^3 − 4 x 5 x^2
x^2 − 5 x + 6 x^2 − 4 x + 4
10 x^3 6 x + 12
- Simplify, cancelling any common factors:
(a)
x^3 − 6 x^2 + 9x x^2 − 6 x + 8 x^2 − 8 x + 17 x − 4
(b)
x^2 − x − 2 5 x^4 − 15 x^3
÷
x^2 + 5x + 4 x^3 + 4x^2
(c)
x^2 − 2 xy x^3 + x^2 y x^2 − 4 y^2 x^2 − 4 xy − 5 y^2
(d)
x
x^2 1 −
x^2
(e)
x − 1
x + 1 1 +
x^2 − 1
(f)
x^3 − 5 x^2 x^2 − 2 x − 15 x^2 + 2x x^2 − 4
(g)
x 1 −
x
x x − 1
- Combine the fractions into a single fraction and simplify:
(a)
x − 2
3 x x^2 − 2 x
x
(b)
x^2 − 5 x + 6
x^2 − 9
(c) 2 −
x
x + 1
(d) x x + y
y x − y
(e) x − 2 y x^2 − xy
2 y x^2 − y^2
(f) x −
x
x + 1 x − 1
- (a) Solve
x + 3
7 x + 1 x + 3
(b) Solve 16 +
3 x x − 4
x − 4
(c) Solve
x^2 − 2 x
x^2 − 4
x^2 + 2x
(d) Solve
x^2 x − 5
x − 5
- Pipe A can fill a tank in 3 hours. If pipes A and B work together, they can fill 5 tanks in 6 hours. How long would it take for pipe B to fill a tank by itself?
- Calvin can eat 800 Brussels sprouts in 4 hours. Phoebe can eat 600 Brussels sprouts in 5 hours. It takes Bonzo 3 hours to eat 240 Brussels sprouts. How long will it take them to eat 600 Brussels sprouts if they work together?
- Calvin can eat 396 tacos in 6 hours. Calvin and Phoebe, eating together, can eat 324 tacos in 3 hours. Phoebe and Bonzo, eating together, can eat 456 tacos in 4 hours. How long will it take Bonzo to eat 576 tacos by himself?
- The numerator of a fraction is 8 less than the denominator. If 7 is added to the top and the bottom of
the fraction, the new fraction is equal to
. Find the original fraction.
- A river flows at a constant rate of 1.6 miles per hour. Calvin rows 32 miles downstream (with the current) in 5 hours less than it takes him to row the same distance upstream (against the current). What is Calvin’s rowing speed in still water?
- Given that f (x) =
x^2 + 2 and g(x) = x^2 , find:
(a) f (3).
(b) f (−1).
(c) f (x + 1).
Thus,
a b c
is not in general equal to
a b c
- Simplify, cancelling any common factors:
(a)
x^2 − x − 2 x^2 + 3x + 2
x^2 − 5 x + 6 x^2 − 4 x + 4
x^2 − x − 2 x^2 + 3x + 2
x^2 − 5 x + 6 x^2 − 4 x + 4
(x − 2)(x + 1) (x + 1)(x + 2)
(x − 2)(x − 3) (x − 2)^2
x − 3 x + 2
(b) x^3 − 4 x 5 x^2
x^2 − 5 x + 6 x^2 − 4 x + 4
10 x^3 6 x + 12
x^3 − 4 x 5 x^2
x^2 − 5 x + 6 x^2 − 4 x + 4
10 x^3 6 x + 12
x(x^2 − 4) 5 x^2
(x − 2)(x − 3) (x − 2)^2
10 x^3 6(x + 2)
x(x − 2)(x + 2) 5 x^2
(x − 2)(x − 3) (x − 2)^2
10 x^3 6(x + 2)
x^2 (x − 3) 3
- Simplify, cancelling any common factors:
(a)
x^3 − 6 x^2 + 9x x^2 − 6 x + 8 x^2 − 8 x + 17 x − 4
x^3 − 6 x^2 + 9x x^2 − 6 x + 8 x^2 − 8 x + 17 x − 4
x(x − 3)^2 (x − 2)(x − 4) x^2 − 8 x + 17 x − 4
x − 4 x − 4
x(x − 3)^2 (x − 2)(x − 4) x^2 − 8 x + 17 + 2(x − 4) x − 4
x(x − 3)^2 (x − 2)(x − 4) x^2 − 6 x + 9 x − 4
x(x − 3)^2 (x − 2)(x − 4) (x − 3)^2 x − 4
x(x − 3)^2 (x − 2)(x − 4)
x − 4 (x − 3)^2
x x − 2
(b)
x^2 − x − 2 5 x^4 − 15 x^3
÷
x^2 + 5x + 4 x^3 + 4x^2
x^2 − x − 2 5 x^4 − 15 x^3
÷
x^2 + 5x + 4 x^3 + 4x^2
x^2 − x − 2 5 x^4 − 15 x^3 x^2 + 5x + 4 x^3 + 4x^2
x^2 − x − 2 5 x^4 − 15 x^3
x^3 + 4x^2 x^2 + 5x + 4
(x − 2)(x + 1) 5 x^3 (x − 3)
x^2 (x + 4) (x + 1)(x + 4)
x − 2 5 x(x − 3)
(c)
x^2 − 2 xy x^3 + x^2 y x^2 − 4 y^2 x^2 − 4 xy − 5 y^2
x^2 − 2 xy x^3 + x^2 y x^2 − 4 y^2 x^2 − 4 xy − 5 y^2
x^2 − 2 xy x^3 + x^2 y
x^2 − 4 xy − 5 y^2 x^2 − 4 y^2
x(x − 2 y) x^2 (x + y)
(x − 5 y)(x + y) (x − 2 y)(x + 2y)
x − 5 y x(x + 2y)
(d)
x
x^2 1 −
x^2
x
x^2 1 −
x^2
x
x^2 1 −
x^2
x^2 x^2
x^2 − 2 x − 3 x^2 − 9
(x − 3)(x + 1) (x − 3)(x + 3)
x + 1 x + 3
(e)
x − 1
x + 1 1 +
x^2 − 1
x − 1
x + 1 1 +
x^2 − 1
x − 1
x + 1 1 +
(x − 1)(x + 1)
x − 1
x + 1 1 +
(x − 1)(x + 1)
(x − 1)(x + 1) (x − 1)(x + 1)
(x − 1)(x + 1) x − 1
(x − 1)(x + 1) x + 1 (x − 1)(x + 1) +
(x − 1)(x + 1) (x − 1)(x + 1)
(x + 1) + (x − 1) (x − 1)(x + 1) + 1
2 x (x^2 − 1) + 1
2 x x^2
x
(f)
x^3 − 5 x^2 x^2 − 2 x − 15 x^2 + 2x x^2 − 4
x^3 − 5 x^2 x^2 − 2 x − 15 x^2 + 2x x^2 − 4
x^3 − 5 x^2 x^2 − 2 x − 15
x^2 − 4 x^2 + 2x
x^2 (x − 5) (x − 5)(x + 3)
(x − 2)(x + 2) x(x + 2)
x(x − 2) x + 3
(g)
x 1 −
x
x x − 1
x 1 −
x
x x − 1
x 1 −
x
x x
x x − 1
x + 1 x − 1
x x − 1
x + 1 − x x − 1
x − 1
- Combine the fractions into a single fraction and simplify:
(a)
x − 2
3 x x^2 − 2 x
x
x − 2
3 x x^2 − 2 x
x
x − 2
3 x x(x − 2)
x
x x
x − 2
3 x x(x − 2)
x − 2 x − 2
x
2 x + 3x − 2(x − 2) x(x − 2
3 x + 4 x(x − 2)
- (a) Solve
x + 3
7 x + 1 x + 3
x + 3
7 x + 1 x + 3
5(x + 3) ·
x + 3
= 5(x + 3) ·
7 x + 1 x + 3
5(x + 3) ·
x + 3
= 5(x + 3) ·
7 x + 1 x + 3 15 + 12(x + 3) = 5(7x + 1) 15 + 12x + 36 = 35x + 5 51 + 12x = 35x + 5 46 + 12x = 35x 46 = 23x 2 = x
Check:
x + 3
7 x + 1 x + 3
The solution is x = 2.
(b) Solve 16 + 3 x x − 4
x − 4
3 x x − 4
x − 4
(x − 4) ·
3 x x − 4
= (x − 4) ·
x − 4
(x − 4) · 16 + (x − 4) ·
3 x x − 4 = (x − 4) ·
x − 4 16(x − 4) + 3x = 12 16 x − 64 + 3x = 12 19 x − 64 = 12 19 x = 76 x = 4
Plugging x = 4 into 16 +
3 x x − 4
x − 4
results in division by zero. Therefore, there are no solutions.
(c) Solve
x^2 − 2 x
x^2 − 4
x^2 + 2x
x^2 − 2 x
x^2 − 4
x^2 + 2x
x(x − 2)
(x − 2)(x + 2)
x(x + 2)
x(x − 2)(x + 2) ·
x(x − 2)
(x − 2)(x + 2)
x(x + 2)
= x(x − 2)(x + 2) · 0
3 x(x − 2)(x + 2) x(x − 2)
x(x − 2)(x + 2) (x − 2)(x + 2)
x(x − 2)(x + 2) x(x + 2)
3(x + 2) − x + (x − 2) = 0 3 x + 6 − x + x − 2 = 0 3 x + 4 = 0 3 x = − 4
x = −
Check: When x = −
x^2 − 2 x
x^2 − 4
x^2 + 2x
The solution is x = −
(d) Solve x^2 x − 5
x − 5
x^2 x − 5
x − 5
(x − 5) ·
x^2 x − 5
= (x − 5) · 7 + (x − 5) ·
x − 5 x^2 = 7(x − 5) + 25 x^2 = 7x − 35 + 25 x^2 = 7x − 10 x^2 − 7 x + 10 = 0 (x − 2)(x − 5) = 0 x = 2 or x = 5
Check: When x = 2, x^2 x − 5
and 7 +
x − 5
However, x = 5 causes division by 0 in the original equation. Therefore, the only solution is x = 2.
- Pipe A can fill a tank in 3 hours. If pipes A and B work together, they can fill 5 tanks in 6 hours. How long would it take for pipe B to fill a tank by itself?
Let x be Calvin’s rate in tacos per hour, let y be Phoebe’s rate in tacos per hour, and let z be Bonzo’s rate in tacos per hour.
hours · tacos per hour = tacos Calvin 6 · x = 396 Calvin and Phoebe 3 · x + y = 324 Phoebe and Bonzo 4 · y + z = 456 Bonzo t · z = 576
The first row says 6x = 396, so x = 66. The second row says 3(x + y) = 324. Plug in x = 66 and solve for y:
3(66 + y) = 324 66 + y = 108 y = 42
The third row says 4(y + z) = 456. Plug in y = 42 and solve for z:
4(42 + z) = 456 42 + z = 114 z = 72
The fourth row says tz = 576. Plug in z = 72 and solve for t:
72 t = 576 t = 8
It takes Bonzo 8 hours.
- The numerator of a fraction is 8 less than the denominator. If 7 is added to the top and the bottom of
the fraction, the new fraction is equal to
. Find the original fraction.
Let n d
be the original fraction. The numerator is 8 less than the denominator, so
n = d − 8.
If 7 is added to the top and the bottom, the result equals
n + 7 d + 7
Plug n = d − 8 into
n + 7 d + 7
d − 8 + 7 d − 7
d − 1 d + 7
Multiply by 5(d + 7) to clear denominators:
5(d + 7) ·
d − 1 d + 7
= 5(d + 7) ·
, 5(d − 1) = 3(d + 7).
Solve for d: 5 d − 5 = 3d + 21, 2 d = 26, d = 13.
Hence, n = d − 8 = 13 − 8 = 5. The fraction is
- A river flows at a constant rate of 1.6 miles per hour. Calvin rows 32 miles downstream (with the current) in 5 hours less than it takes him to row the same distance upstream (against the current). What is Calvin’s rowing speed in still water?
Let x be Calvin’s rowing speed in still water. His downstream rate is x + 1.6, while his downstream rate is x − 1 .6. Let t be the time it takes Calvin to row upstream. Then it takes him t − 5 (that is, 5 hours less) to row downstream. time · speed = distance downstream t − 5 · x + 16 = 32 upstream t · x − 1. 6 = 32
The equations are (t − 5)(x + 1.6) = 32 and t(x − 1 .6) = 32. Solve the second equation for t:
t(x − 1 .6) = 32 1 x − 1. 6
· t(x − 1 .6) =
x − 1. 6
t =
x − 1. 6
Plug this into (t − 5)(x + 1.6) = 32 and solve for x:
(t − 5)(x + 1.6) = 32 ( 32 x − 1. 6
(x + 1.6) = 32
(x − 1 .6) ·
x − 1. 6
(x + 1.6) = (x − 1 .6) · 32 ( 32(x − 1 .6) x − 1. 6
− 5(x − 1 .6)
(x + 1.6) = 32(x − 1 .6)
[32 − 5(x − 1 .6)](x + 1.6) = 32(x − 1 .6) (32 − 5 x + 8)(x + 1.6) = 32x − 51. 2 (40 − 5 x)(x + 1.6) = 32x − 51. 2 − 5 x^2 + 32x + 64 = 32x − 51. 2 − 5 x^2 = − 115. 2 x^2 = 23. 04 x = ± 4. 8
Since Calvin’s speed can’t be negative, I must have x = 4.8 miles per hour.
- Given that f (x) =
x^2 + 2
and g(x) = x^2 , find:
Since f (x) =
x + 1 x^2 − 3 x + 2
x + 1 (x − 1)(x − 2)
, f will be undefined if x = 1 or if x = 2. The domain
consists of all real numbers except for 1 and 2.
(b) Find the domain of g(x) =
x − 1.
g will be undefined if the stuff inside the square root (x − 1) is negative. So I must have x − 1 ≥ 0, or x ≥ 1. The domain consists of all real numbers greater than or equal to 1.
- Find the range of the function whose entire graph is pictured below.
The range of a function is the set of its outputs, i.e. the set of y-values it attains. The function attains every y-value (height) from −2 to 3, including 3 but not including −2 (the open circle at (− 2 , −2) indicates that the point is not on the graph). Therefore, the range is − 2 < y ≤ 3.
- (a) y varies directly with x. When x = 5, y = 135. Find x when y = 9.
y varies directly with x: y = kx. When x = 5, y = 135: 135 = k · 5, so k = 27. Therefore, y = 27x. When y = 9, I get 9 = 27x, so x =
(b) y is inversely proportional to x. When x = 8, y = 24. Find x when y = 16.
y is inversely proportional to x: y =
k x
When x = 8, y = 24: 24 = k 8
, so k = 8 · 24 = 192. Therefore, y =
x
When y = 16, I get 16 =
x
. Multiplying both sides by x, I get 16x = 192, so x =
(c) y is inversely proportional to the square of x. When x = 6, y = 20. Find y when x = 4.
y is inversely proportional to the square of x: y =
k x^2
When x = 6, y = 20: 20 = k 62
, or 20 = k 36
. Then k = 20 · 36 = 720, so y =
x^2
When x = 4, y =
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©^ c2008 by Bruce Ikenaga 13
