Chapter 8.4, Problem 36PPSE

Solution

a.

To calculate:A simplified expression for the area of the rectangle for the figure given below.

  

The area of the rectangle is $\frac{6\left(a+1\right)}{a-3}$ square units.

Given information:

The rectangle as given below.

  

Formula Used:

To simplify the ration expressions, identify the restrictions of the function.

To multiply the rational expressions, look for the numerator and denominator to see if there are any common factors that can be cancelled out.

To divide the rational expressions, multiply the dividend by the reciprocal of the divisor.

Calculation:

Consider the rectangle as given below,

  

Multiply the length and width to get the area of rectangle.

  $A=\frac{4a+4}{a+3}\cdot \frac{3a+9}{2a-6}$

Take out common factor.

  $A=\frac{4\left(a+1\right)}{a+3}\cdot \frac{3\left(a+3\right)}{2\left(a-3\right)}$

Cancel out the common factor and simplify.

  $A=\frac{6\left(a+1\right)}{a-3}$

Thus, the area of the rectangle is $\frac{6\left(a+1\right)}{a-3}$ square units.

b.

To identify:The parts of the expression to analyse to determine the restriction on $a$ .

To put the restrictions on $a$ , analyse the denominator of the length and width of the rectangle.

Given information:

The rectangle as given below.

  

Explanation:

Consider the rectangle as given below,

  

To put the restrictions on $a$ , analyse the denominator of the length $\frac{4a+4}{a+3}$ and width $\frac{3a+9}{2a-6}$ .

The length and width of the rectangle will be invalid on the value of $a$ where $a=0$ .

c.

To state:All the restriction on $a$ .

To put the restrictions on $a$ , analyse the denominator of the area.

Given information:

The rectangle as given below.

  

Explanation:

Consider the rectangle as given below,

  

To put the restrictions on $a$ , analyse the denominator of the length $\frac{4a+4}{a+3}$ and width $\frac{3a+9}{2a-6}$ .

The length and width of the rectangle will be invalid on the value of $a$ where $a=0$ .

  $\begin{array}{c}a+3=0\\ a=-3\end{array}$

  $\begin{array}{c}2a-6=0\\ 2a=6\\ a=3\end{array}$

Thus, all the restriction on $a$ are $a\ne -3,3$ .