# Two planes which are 3720 miles apart, fly toward each other. Their speeds differ by 30 mph. If they pass each other in 4 hours, what is the speed of each?

Jun 14, 2017

$480 m p h \mathmr{and} 450 m p h$

#### Explanation:

let say thier speed are ${v}_{1}$ and ${v}_{2}$ respectively.
therefore,
${v}_{1} - {v}_{2} = 30 \to i$ and

${v}_{1} t + {v}_{2} t = 3720$

$t \left({v}_{1} + {v}_{2}\right) = 3720$

since $t = 4$,
${v}_{1} + {v}_{2} = \frac{3720}{4} = 930 \to i i$

we can find ${v}_{1}$ and ${v}_{2}$ by solving silmutaneos equations $i$ and $i i$

let say we use eliminate method $\left(i + i i\right)$
$2 {v}_{1} = 960$

${v}_{1} = \frac{960}{2} = 480$ mph

replace ${v}_{1} = 480$ into $i$,

$480 - {v}_{2} = 30$

${v}_{2} = 450$ mph