Question

Write an absolute value inequality that represents the situation.
Then solve the inequality.

The difference between the perimeters of the figures is less than or equal to $3$.

Answer 1

Illustration of figures:

Considering a rectangle of sides $x+1$ and $3$ and a square of side $x$

Now perimeter of rectangle $=\left[2\right(3)+2(x+1\left)\right]$

Now perimeter of square $=4x$

According to the question the equation of inequality

$\Rightarrow \left[2\right(3)+2(x+1\left)\right]-4x\le 3$

Solving the inequality

$$\begin{gathered} \Rightarrow 6+2x+2-4x\le 3 \\ \Rightarrow 8-2x\le 3 \\ \Rightarrow 8-2x-3\le 0 \\ \Rightarrow 5-2x\le 0 \\ \Rightarrow 5\le 2x \\ \Rightarrow \frac{5}{2}\le x \\ \Rightarrow x\ge \frac{5}{2} \end{gathered}$$

Hence, the solution of inequality is $x\ge \frac{5}{2}$.