Chapter 6.2, Problem 32PPE

To determine

To write: The system of linear equations for the given scenario and solve it to obtain dimensions of rectangle.

Answer to Problem 32PPE

l = 13 and w = 4

Explanation of Solution

Given information:

Following rectangle is given:

  

Perimeter of rectangle = 34 cm

The length ( l ) of rectangle is 5 more than twice the width ( w ).

Calculation:

Based on given information

  $2\left(l+w\right)=34\left(\because perimeter=2\left(lenght+breadth\right)\right)$

Also $l=5+2w$ as length ( l ) of rectangle is 5 more than twice the width ( w ).

Thus system of equations for the given scenarios is as follows:

  $2l+2w=34$

  $l=5+2w$

Let’s solve above two equation using substitution method.

Substitute $5+2w$ for l in equation $2l+2w=34$

  $2\left(5+2w\right)+2w=34$

Now simplify above equation.

  $\begin{array}{l}2\left(5+2w\right)+2w=34\\ \Rightarrow 10+4w+2w=34\\ \Rightarrow 6w=34-10\\ \Rightarrow 6w=24\\ \Rightarrow w=4\end{array}$

Use above value ofw in equation $l=5+2w$

  $\begin{array}{c}l=5+2\left(4\right)\\ =5+8\\ =13\end{array}$

Therefore the solution is l = 13 and w = 4.

Check

Area of rectangle = $l\times w=13\times 4=52c{m}^2$ that is same as provided in figure.