## Precalculus (6th Edition) Blitzer

The solution set of the equation $6\left| 1-2x \right|-7=11$ is $\left\{ -1,2 \right\}$.
Let us consider the following equation involving the absolute value, $6\left| 1-2x \right|-7=11$ Solve the equation as given below: Add 7 to both sides: \begin{align} & 6\left| 1-2x \right|-7+7=11+7 \\ & 6\left| 1-2x \right|=18 \end{align} Divide by 6 on both sides, \begin{align} & \frac{6\left| 1-2x \right|}{6}=\frac{18}{6} \\ & \left| 1-2x \right|=3 \end{align} By using the definition of an absolute function, $\left| x \right|=a\Rightarrow x=a\text{ or }x=-a$ \begin{align} & \left| 1-2x \right|=3 \\ & 1-2x=3\text{ or 1}-2x=-3 \end{align} Solve the equation $1-2x=3$. Subtract 1 from both sides: \begin{align} & 1-2x-1=3-1 \\ & -2x=2 \end{align} And divide by $-2$ on both sides, \begin{align} & \frac{\left( -2x \right)}{\left( -2 \right)}=\frac{2}{\left( -2 \right)} \\ & x=-1 \end{align} Then, solve the second part of the equation $1-2x=-3$ Subtract 1 from both sides, \begin{align} & 1-2x-1=-3-1 \\ & -2x=-4 \end{align} Divide by $-2$ on both sides, \begin{align} & \frac{\left( -2x \right)}{\left( -2 \right)}=\frac{\left( -4 \right)}{\left( -2 \right)} \\ & x=2 \end{align} Therefore, $x=-1$ is the solution of the equation. Check $x=-1$ in the original equation, \begin{align} & 6\left| 1-2\times \left( -1 \right) \right|-7=11 \\ & 6\left| 1+2 \right|-7=11 \\ & 6\left| 3 \right|-7=11 \\ & 6\times 3-7=11 \end{align} And simplify further, \begin{align} & 18-7=11 \\ & 11=11\text{ }\left( \text{true} \right) \end{align} Therefore, $x=-1$ is the solution of the equation. Thus, the solution set of the equation $6\left| 1-2x \right|-7=11$ is $\left\{ -1,2 \right\}$.