Chapter 2.4, Problem 31P

A square and a rectangle have the same perimeters. The length of a side of the square is $4x-1.$

The length of the rectangle is $2x+1$ and the width is $x+2$

Write and solve an equation to find $x$.

To determine

To write and solve an equation to find $x$ for given problem “a square and a rectangle have the same perimeters. The length of a side of the square is $4x-1$ . The length of the rectangle is $2x+1$ and the width is $x+2$ ”

Answer to Problem 31P

  $4\left(4x-1\right)=2\left(3x+3\right)$ is the required equation and the value of $x$ is $1$ .

Explanation of Solution

Given information :

Given “ a square and a rectangle have the same perimeters. The length of a side of the square is $4x-1$ . The length of the rectangle is $2x+1$ and the width is $x+2$ ”

Given length of of a side of the square is $4x-1$ and

the length of the rectangle is $2x+1$ and the width is $x+2$ ”

Thus, perimeter of a square is $4\left(4x-1\right)$ ............. $(i)$

perimeter of a rectangle is $2\left(2x+1+x+2\right)=2\left(3x+3\right)$ ............. $(ii)$

From given condition, we write $(i)=(ii)$

Thus, $4\left(4x-1\right)=2\left(3x+3\right)$ is the required equation for the given statement.

Now to solve $4\left(4x-1\right)=2\left(3x+3\right)$

  $\begin{array}{l}\Rightarrow 2\left(4x-1\right)=\left(3x+3\right)\\ \Rightarrow 8x-2=3x+3\\ \Rightarrow 8x-3x=3+2\\ \Rightarrow 5x=5\\ \therefore x=1\end{array}$

Hence, $4\left(4x-1\right)=2\left(3x+3\right)$ is the required equation and the value of $x$ is $1$ .