## Precalculus (6th Edition) Blitzer

The width of the field is $50$ yards and the length is $100$ yards.
Let us assume the width of the field is $x$. Then, the length would be $x+50$. And equate the perimeter of the field as: \begin{align} & 2\left[ x+\left( x+50 \right) \right]=300 \\ & 2\left[ 2x+50 \right]=300 \\ & 2x+50=150 \\ & 2x=150-50 \end{align} Then, simplify it further: \begin{align} & x=\frac{100}{2} \\ & x=50 \end{align} Therefore, the width of the field is $50$ and the length is $50+50=100$. Thus, the width of the field is $50$ and the length is $100$.