Q51. - Algebra 2

Question

Solve each inequality. Graph the solution set on a number line.

$\left|\mathbf{3}\mathbf{b}\mathbf{+}\mathbf{11}\right|>\mathbf{1}$

Step-by-Step Solution

Answer

Solution set of inequality is $\left\{\mathbf{b}\mathbf{|}\mathbf{-}\frac{\mathbf{10}}{\mathbf{3}}<\mathbf{b}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{b}<\mathbf{-}\mathbf{4}\right\}$.

Solution set on number line is 

Step 1 - Write properties of inequality.

Properties of inequalities are written in following table:

Step 2 - Solve 3 b + 11 &#62; 1 .

$\left|3b+11\right|>1$ is equivalent to $3b+11>1$ or $3b+11<-1$.

Consider, $3b+11>1$. Subtract 11 from each part then divide each part by 3 and simplify.

$\begin{array}{c}3b+11-11>1-11\\ \frac{3b}{3}>\frac{-10}{3}\\ b>\frac{-10}{3}\end{array}$

Step 3 &ndash; Description of step.

Consider, $3b+11<-1$. Subtract 11 from each part then divide each part by 3 and simplify.

 $\begin{array}{c}3b+11-11<-1-11\\ \frac{3b}{3}<\frac{-12}{3}\\ b<-4\end{array}$

Thus, solution is $b>\frac{-10}{3}$ or $b<-4$.

Step 4 &ndash; Graph the solution set.

As solution set of the inequality is all real numbers greater than $-\frac{10}{3}$ or less than $-4$.

The solution set is, $\left\{b|-\frac{10}{3}<b\text{ or }b<-4\right\}$.

Open circles on $-\frac{10}{3}$ and $-4$ shows these points are excluded in the solution set.