## Precalculus (6th Edition) Blitzer

The force required is $20$ pounds.
Consider that the distance by which the spring will stretch is D and the force required to stretch the spring is F. It is provided that the variable D varies directly as F, that is $D\ =\ kF$. Thus, \begin{align} & 9=12k \\ & k=\frac{9}{12} \\ & k=\frac{3}{4} \end{align} Substitute D = 15 and $k=\frac{3}{4}$ in $D\ =\ kF$. \begin{align} & D=kF \\ & 15=\frac{3}{4}F \\ & F=\frac{15\times 4}{3} \\ & F=20 \end{align} Thus, a force of 20 pounds is required to stretch the spring by 15 inches.