Section 6.1: Exercises 102, and 120
Which numbers cannot be used in place of the variable in each rational expression?

− 2x − 8 102. x + 2 x − 8
2

=

−2(x + 4) (x + 4)(x − 2) =− 2 x−2

120 . Given C(p) = (a) (b)

From the graph, we have C(90) = $50,000 C(95) = $100,000 Plug in p = 99.5, we get C(99.9) = C(99.5) = 500000 500000 = = $1,000,000 100 − 99.5 0.5

500000 100−p

Plug in p = 99.9, we get

c)

As p = 100%, the cost would be undefined.

500000 500000 = = $50,000,000 0.01 100 − 99.9

Section 6.2: Exercises 76
76. (w − 1) w2 − 1
2

×

w −1 w + 2w + 1
2

Using the formulae w 2 + 2w + 1 = (w + 1)2 and w 2 − 1 = (w − 1)(w + 1)

=

(w − 1)(w + 1) w−1 × (w − 1)2 (w + 1)2 = 1 w+1

Section 6.4: Exercises 66 and 90
66. Add:

2x 3x + 2 . x − 9 x + 4x + 3
2

3x 2x + 2 x 2 − 9 x + 4x + 3 2x 3x = + (x − 3)(x + 3) (x + 1)(x + 3) 2x(x + 1) + 3x(x − 3) = (x + 1)(x − 3)(x + 3) 2x 2 + 2x + 3x 2 − 9x = (x + 1)(x − 3)(x + 3) 5x 2 − 7x = (x + 1)(x − 3)(x + 3)

One day work by Melanie =

One day work by Melissa = 2x Therefore one day work by working together

1

x 1

Plug inx = 5, we get

=

Section 6.5: Exercises 58
Simplify each complex fraction. Reduce each answer to lowest terms.

3 3 = 10 2∗5

1 1 + x 2x 3 = 2x

ab + b 2 a + b / 2 4 5 6a b 58. 4ab
Numerator = ab+b2 4ab5

=

b(a+b) 4ab5

=

4ab4

a+b

Denominator = So

6a2 b4

a+b

a + b 6a2 b4 = ∗ 4ab 4 a + b = 3a 2

a+b 4 = 4ab a+b 6a2 b 4

ab + b2 4ab 5 a+b 6a2 b 4

Section 6.6 Exercises 52 and 68 Solve each equation. 52.
Multiplying by 2a a+4 a+4 = a 2

Using zero product property

a(a + 4) − 2(a + 4) = 0 a + 4 = 0 or a − 2 = 0 a = −4 or a = 2 (a + 4)(a − 2) = 0

a(a + 4) = 2(a + 4)

We have, the formula

1 1 1 = + f o i

Now given �� = 250 mm, o = 2,000,000 mm. Using these, in the above formula, we get 1 1 1 = − i 250 2000000 1 8000 − 1 = i 2000000 1 7999 = i 2000000 2000000 7999

1 1 1 = − i f o

Cross multiplying:

i = 250.03 mm

i =

Section 6.7 Exercises 32 and 62
Solve each proportion.

32. Solve

9 3 = x 2
9 3 = x 2

Cross multiply Dividing by 3

3x = 2(9) 3x = 18 x=6

Cross multiplying

2x = 50 ∗ 100 2x = 5,000

x 100 = 50 2

5,000 2 x = 2,500 is the estimated population of bears. Dividing by 2 x=

Section 6.8 Exercises 38 and 54
38. Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one-half hour longer than Smith’s. How fast was each one traveling?

Let Smith’s goes x miles per hour then Jones goes x + 5 miles per hour. 45 Then time taken by Smith to go 45 miles = And the time taken by Jones to go 75 miles = Now since Jones took one-half hour longer than Smith’s, so we have 75 45 1 = + x+5 x 2 75 45 1 − = x+5 x 2 x x+5 75

75x 45(x + 5) 1 − = x(x + 5) x(x + 5) 2 75x − 45x − 225 1 = x(x + 5) 2 60x − 450 = x 2 + 5x 30x − 225 1 = x 2 + 5x 2 75x − 45(x + 5) 1 = x(x + 5) 2

x 2 + 5x − 60x + 450 = 0

2(30x − 225) = x 2 + 5x

x(x − 45) − 10(x − 45) = 0 Using zero product property x − 45 = 0 or x − 10 = 0 x = 45 or x = 10 (x − 45)(x − 10) = 0

x 2 − 45x − 10x + 450 = 0

x 2 − 55x + 450 = 0

Hence Smith’s goes 45 miles per hour and Jones goes 50 miles per hour or Hence Smith’s goes 10 miles per hour and Jones goes 15 miles per hour

speed of the plane in the opposite direction to the wind = (180 – v) mph, therefore the 5 time taken to cover 5 mile against the wind = . It is given that the plane flies 7 miles 5 7 = 180 + �� 180 − ��
180−��

Let speed of the wind be v mph. Then speed of the plane in the direction of the wind 7 . Now the = (180 + ��) mph therefore the time taken to cover 7 mile with the wind =
180+��

with the wind in the same amount of time as it flies 5 miles against the wind, therefore we have

Cross multiply Using distributive property

5(180 + ��) = 7(180 − ��) 900 + 5�� = 1260 − 7�� 5�� + 7�� = 1260 − 900

Collecting like terms in one side

Dividing by 12, we obtain

12�� = 360 �� = 360 12 �� = 30

Hence speed of the wind is 30 mph