Chapter 9.5, Problem 31P

To determine

To find: The original dimension of the rectangle by using variable.

Answer to Problem 31P

The original length of the rectangle is $30\text{unit}$ , and the original width of the rectangle is $28\text{unit}$ .

Explanation of Solution

Given information:

The length of a rectangle is increased by $12$ .

The width of the rectangle is decreased by $8$ .

The original length is increased by $5$ and width decreased by $4$ .

The area is unchanged.

Calculation:

Let, original length $=x$ , and original width $=y$ .

  $\text{Original}\text{area}=xy$ .

If the length is increased by $12$ and width is decreased by $8$ , area is unchanged.

  $\begin{array}{c}\left(x+12\right)\left(y-8\right)=xy\\ xy-8x+12y-96=xy\\ 12y-8x=96\end{array}$ … (1)

If the length is increased by $5$ , and width is decreased by $4$ , area is unchanged.

  $\begin{array}{c}\left(x+5\right)\left(y-4\right)=xy\\ xy-4x+5y-20=xy\\ 5y-4x=20\end{array}$ … (2)

Multiply by $2$ in equation $\left(2\right)$ .

  $\begin{array}{c}2\left(5y-4x\right)=2\left(20\right)\\ 10y-8x=40\end{array}$ … (3)

Subtract equation $\left(3\right)$ from equation $\left(1\right)$ .

  $\begin{array}{c}12y-8x=96\\ 10y-x=40\end{array}$

  $2y\text{ =56}$

Solve the result,

  $y=28$

Substitute the value of $y$ in equation $\left(2\right)$ .

  $\begin{array}{c}5\left(28\right)-4x=20\\ 140-4x=20\\ x=30\end{array}$

Therefore,

  $\begin{array}{l}x=30\\ y=28\end{array}$

Therefore, the original length of the rectangle is $30\text{unit}$ , and the original width of the rectangle is $28\text{unit}$ .