Q24. - Algebra 1

Question

The perimeter of a rectangle is 48 inches. The length is 6 inches greater than the width. Define the variables, and write equations to represent this situation. Solve the system by using substitution.

Step-by-Step Solution

Answer

The solution of the system of equations is 

 $\begin{array}{l}\mathbf{l}\mathbf{=}\mathbf{15}\mathbf{\text{\hspace{0.17em}inches}}\\ \mathbf{w}\mathbf{=}\mathbf{9}\mathbf{\text{\hspace{0.17em}inches}}\end{array}$ 

Where length of a rectangle is $l$ and width of a rectangle is $w$.

Step 1. State the concept of substitution method to solve system of equations.

The substitution method is used to find the set of solutions for the given system of equations. It is the method in which one variable is first solved in the form of another variable. 

So, the value of one variable in the form second variable is put on the second equation. It will give the solution of the one variable. And then solve it for another variable.

Step 2. Solve at least one equation for one variable and substitute.

Let length of a rectangle be $l$ and width of a rectangle be $w$, then as per conditions of problem the given system of equation is:

 $2l+2w=48$                       … (1)

$l=6+w$                              … (2)

Note that in equation (2), variable $l$ is in terms of $w$, therefore, put value of $l$ in eq. (1):

$\begin{array}{c}2l+2w=48\\ 2(6+w)+2w=48\\ 12+2w+2w=48\\ 12+4w=48\\ 4w=36\\ w=9\end{array}$ 

Step 3. Solve for l .

Substitute 8 for $w$ into the equation (2) and solve for $l$.

  $\begin{array}{c}l=6+w\\ =6+9\\ =15\end{array}$

Therefore, the length and width of rectangle is calculated as: 

 $\begin{array}{l}l=15\text{\hspace{0.17em}inches}\\ w=9\text{\hspace{0.17em}inches}\end{array}$