Question

n a particular day, the wind added 2 miles per hour to Alfonso's rate when he was cycling with the wind and subtracted 2 miles per hour from his rate on his return trip. Alfonso found that in the same amount of time he could cycle 63 miles with the wind, he could go only 51 miles against the wind.What is his normal bicycling speed with no wind?

Answer 1

Recall your d = rt, or distance = rate * time

notice, the time he took to cover 63 miles with the wind, is the same amount of time he took against it with 51 miles only, let's say, he took "t" hours long
and he was cycling at a rate of "r"

thus    $\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{with the wind}&63&r+2&t\\ \textit{against the wind}&51&r-2&t \end{array}\\\\ -----------------------------\\\\ \begin{cases} 63=(r+2)t\implies \cfrac{63}{r+2}=\boxed{t}\\\\ 51=(r-2)t\\ ----------\\ 51=(r-2)\left( \boxed{\cfrac{63}{r+2}} \right) \end{cases}$

solve for "r"