91Ó°ÊÓ

The value of the perimeter of a rectangle is 32 inches and the value of the area of the rectangle is 63 square inches.

Let $x$be the length of the rectangle and $y$be the width of the rectangle.

Then perimeter of the rectangle will be $2\left(x+y\right)$and the area of the rectangle will be $xy$. According to the given conditions, we have the following equations:

$$\begin{gathered} 2\left(x+y\right)=32--->\left(1\right) \\ xy=63--->\left(2\right) \end{gathered}$$

From equation (2), solving for $y$gives

$y=\frac{63}{x}$

Substituting this in equation (1) gives

$$\begin{gathered} 2\left(x+\frac{63}{x}\right)=32 \\ x+\frac{63}{x}=16 \end{gathered}$$

Multiplying both sides by x and then simplifying:

$$\begin{gathered} x^2+63=16x \\ {x}^2-16x+63=0 \end{gathered}$$

Factoring the quadratic equation:

$$\begin{gathered} x^2-9x-7x+63=0 \\ \left(x-9\right)\left(x-7\right)=0 \end{gathered}$$

So, either $\left(x-9\right)=0$or $\left(x-7\right)=0$.

When $x-9=0$, then $x=9$ and when $x-7=0$, then $x=7$.

Substitute the values of $x$in the second equation:

When $x=9$:

$y=\frac{63}{9}=7$

When $x=7$, then $y=\frac{63}{7}=9$

Expressing the solution as ordered pairs$\left(x,y\right)$:$\left(9,7\right)\mathrm{or}\left(7,9\right)$