#### Answer

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.