Question

How do you solve $\sqrt{2x+10}-6=2$?

Answer 1

Simplify the equation $\sqrt{2x+10}-6=2$:

Simplifying by adding $6$on both sides:
$\begin{array}{rcl}\sqrt{2x+10}-6+6& =& 2+6\\ & \Rightarrow & \sqrt{2x+10}=2+6\\ & \Rightarrow & \sqrt{2x+10}=8\end{array}$
To remove the radical sign, square both sides as follows:
$\begin{array}{rcl}(\sqrt{2x+10}{)}^2& =& 8^2\\ & \Rightarrow & 2x+10=64\end{array}$
Subtract by 10 on both sides:
$\begin{array}{rcl}2x+10-10& =& 64-10\\ & \Rightarrow & 2x+10-10=64-10\\ & \Rightarrow & 2x=54\end{array}$

Divide by $2$ into both sides:
$\begin{array}{rcl}\frac{2x}{2}& =& \frac{54}{2}\\ & \Rightarrow & x=27\end{array}$

Hence, the value of $x$for the given equation $\sqrt{2x+10}-6=2$ is $27$.