Chapter 9.2, Problem 13CYU

To determine

Tofind: the perimeter of the rectangle.

Answer to Problem 13CYU

The perimeter of the rectangle is $\frac{2x-22}{x^2-x-2}$ .

Explanation of Solution

Given information:

Length of the rectangle is $\frac{4}{x+1}$ and width of the rectangle is $\frac{3}{x-2}$ .

Calculation:to find the perimeter, first use the formula and substitute,

  $\begin{array}{l}p=2\left(l+b\right)\\ p=2\left(\frac{4}{x+1}-\frac{3}{x-2}\right)\text{ [substitute]}\end{array}$

to find the simplified expression, first take LCM of the denominator and then simplify,

LCM of denominator,

  $\begin{array}{l}x+1=\left(x+1\right)\text{ [factor]}\\ x-2=\left(x-2\right)\text{ [factor]}\\ \text{LCM}=\left(x-2\right)\left(x+1\right)\end{array}$

Now, simplify the expression,

  $\begin{array}{l}p=2\left(\frac{4}{x+1}-\frac{3}{x-2}\right)\\ p=2\left(\frac{4\left(x-2\right)-3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\right)\text{ [LCM of denominator and simplify]}\\ p=\frac{2x-22}{x^2-x-2}\text{ [multiply and simplify]}\end{array}$

Thus, the perimeter of rectangle is $\frac{2x-22}{x^2-x-2}$ .