Question 1139823
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The trip going is 4 miles shorter than the trip returning:<br>
let x = length (miles) of trip returning
then x-4 = length of trip going<br>
His speed going is 8mph; his speed returning is 7mph.  So<br>
x/7 = time returning
(x-4)/8 = time going<br>
His time returning is 1 hour more than his time going:<br>
{{{x/7 = (x-4)/8+1}}}<br>
That turns into a linear equation that is easily solved....