## Precalculus (6th Edition) Blitzer

$b=3\ ,l=\ 8$.
We have the formula for the perimeter: $P=2\left( l+b \right)$ So \begin{align} & 22=2\left( l+b \right) \\ & 11=l+b \\ & l=11-b \end{align} The area is $A=l\times b$ , So, $24=l\times b$ Substituting the value of the length in equation $24=l\times b$, we get ${{b}^{2}}-11b+24=0$ On solving the above quadratic equation, we get $b=3\text{ and}\ 8$. Substituting $b$ in equation $l=11-b$. When $b=3$ the value of $l=8$ and when $b=8$ the value of $l=3$.