Q37. - Algebra 1

Question

Solve each inequality. Then graph the solution set.

$\left|2c+3\right|\mathbf{\le }\mathbf{11}$

Step-by-Step Solution

Answer

The solution for the given inequality $\left|2c+3\right|\le 11$ is $\mathit{c}\in \left[-\mathbf{7}\mathbf{,}\mathbf{4}\right]$.

The graph of the solution set which is $c\in \left[-7,4\right]$ is:

Step 1. Solve the given inequality | 2 c + 3 | ≤ 11 .

The solution of the given inequality $\left|2c+3\right|\le 11$ is:

Case 1: $2c+3$ is non-negative.

$\begin{array}{c}2c+3\le 11\\ 2c+3-3\le 11-3\\ 2c\le 8\\ \frac{2c}{2}\le \frac{8}{2}\\ c\le 4\\ c\in \left(-\infty ,4\right]\end{array}$

Case 2: $2c+3$ is negative.

 $\begin{array}{c}-\left(2c+3\right)\le 11\\ \left(-1\right)\left(-\left(2c+3\right)\right)\ge \left(-1\right)\left(11\right)\\ 2c+3\ge -11\\ 2c+3-3\ge -11-3\\ 2c\ge -14\\ \frac{2c}{2}\ge \frac{-14}{2}\\ c\ge -7\\ c\in \left[-7,\infty \right)\end{array}$

The solution of the inequality $\left|x\right|\le a$ is $x\ge -a$ and $x\le a$.

That implies the solution of the inequality $\left|x\right|\le a$ is the intersection of the solutions of the inequalities $x\ge -a$ and $x\le a$.

Find the intersection of the solutions of the inequalities $2c+3\le 11$ and  $2c+3\ge -11$ to find the solution of the inequality $\left|2c+3\right|\le 11$.

The intersection of the solutions of the inequalities $2c+3\le 11$ and $2c+3\ge -11$ is:

$c\in \left[-7,4\right]$

Therefore, the solution of the inequality $\left|2c+3\right|\le 11$ is $c\in \left[-7,4\right]$.

Step 2. Draw the graph of the solution set which is c ∈ [ − 7 , 4 ] .

The graph of the solution set which is $c\in \left[-7,4\right]$ is: