# Chapter 6 - Section 6.7 - Formulas and Applications of Rational Equations - Exercise Set - Page 476: 28

engine 1: $35$ mph engine 2: $40$ mph

#### Work Step by Step

distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$ Let x be the rate of engine 1. The rate of engine 2 is $x+5.$ We are given: total time = $\displaystyle \frac{140}{\text{rate of engine 1}} + \displaystyle \frac{200}{\text{rate of engine 2.}}=9$ That is, $\displaystyle \frac{140}{x}+\frac{200}{x+5} =9\qquad$ ... $/\times x(x+5)$ $140(x+5)+200x=9x(x+5) \qquad$ ... distribute... $140x+700+200x=9x^{2}+45x \qquad$ ... add $-700-340x$ $0=9x^{2}-295x-700$ ... Trying with synthetic division, we find: $$\begin{array}{rrrrrr} 35 & | & 9 & -295 & -700 & \\ & & & 325 & 700 & \\ & & -- & -- & -- & \\ & & 9 & 20 & 0 & \end{array}$$ $0=(x-35)(9x+20)$ x can't be negative, so $\mathrm{x}=35$ mph... the rate of engine 1, so the rate of engine 2 is $40$ mph.

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