(60 / s) + [100 / (s – 10)] = 3

Now I want to combine the two terms on the left side to form one term, you’ll see why later:

(60 / s) + [100 / (s – 10)] = 3

=> [60(s – 10) + 100s] / s(s – 10) = 3

=> [60(s – 10) + 100s] / (s^2 – 10s) = 3

I actually want to get rid of the fraction on the left side, so I’m going to multiply both sides by what’s in the denominator of the fraction:

[60(s – 10) + 100s] / (s^2 – 10s) = 3

=> 60(s – 10) + 100s = 3(s^2 – 10s)

=> 60s – 600 + 100s = 3s^2 – 30s

Combine like terms and simplify:

60s – 600 + 100s = 3s^2 – 30s

=> 3s^2 – 60s – 100s – 30s + 600 = 0

=> 3s^2 – 190s + 600 = 0

The numbers are pretty big, so I’m going to use the quadratic formula to solve for s. Recall: For the equation,

ax^2 + bx + c = 0, the quadratic formula is:

x = [-b +/- √(b^2 – 4ac)] / 2a, so:

3s^2 – 190s + 600 = 0

=> s = [190 +/- √(190^2 – 4(3)(600))] / 2(3)

=> s = [190 +/- √(28,900)] / 6

=> s = (190 +/- 170) / 6

=> s = (190 + 170) / 6 OR (190 – 170) / 6

=> s = 360/6 OR 20/6

=> s = 60 or 10/3

Therefore, s = 60 or 10/3.

(Are you sure your textbook says that 60 is the only answer?

Check the answer carefully because two answers are possible for this problem – even try graphing it to verify.)

( (160 – 10s ) ) / s = 3

160 – 10s = 3s

160 = 13s

s = 160/13