Question

Solve $\left|\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{11}\right|\mathbf{>}\mathbf{1}$ for $\mathbf{x}$.

Short answer

$\mathbf{x}\mathbf{>}\mathbf{-}\mathbf{5}\text{or}\mathbf{x}\mathbf{<}\mathbf{-}\mathbf{6}$.

Step-by-step solution

– Description of step.

For all real numbers $p$ and $q$, $q>0$,

Case1) If $\left|p\right|<q$ then $-q<p<q$.

Case2) If $\left|p\right|>q$ then $p>q\text{or}p<-q$.

– Description of step.

Apply case 2 to $\left|2x+11\right|>1$ which implies $2x+11>1$or $2x+11<-1$.

– Description of step.

Solve $2x+11>1$. Subtract 11 from each side of $2x+11>1$.

$\begin{array}{c}2x+11-11>1-11\\ 2x>-10\end{array}$

– Description of step.

Divide each side of the obtained inequality by 2.

$\begin{array}{c}\frac{2x}{2}>-\frac{10}{2}\\ x>-5\end{array}$

– Description of step.

Solve $2x+11<-1$. Subtract 11 from each side of $2x+11<-1$.

$\begin{array}{c}2x+11-11<-1-11\\ 2x<-12\end{array}$

– Description of step. 

Divide each side of the obtained inequality by 2.

$\begin{array}{c}\frac{2x}{2}<-\frac{12}{2}\\ x<-6\end{array}$

Therefore, $x>-5\text{or}x<-6$.