# How do you solve sqrt(2x+3)=6?

Jun 23, 2018

$x = \frac{33}{2}$

#### Explanation:

$\textcolor{b l u e}{\text{square both sides}}$

${\left(\sqrt{2 x + 3}\right)}^{2} = {6}^{2}$

$2 x + 3 = 36$

$\text{subtract 3 from both sides}$

$2 x = 36 - 3 = 33$

$\text{divide both sides by 2}$

$x = \frac{33}{2}$

$\textcolor{b l u e}{\text{As a check}}$

$\sqrt{2 \times \frac{33}{2} + 3} = \sqrt{33 + 3} = \sqrt{36} = 6 \leftarrow \text{correct}$

Jun 23, 2018

See a solution process below:

#### Explanation:

First, square both sides of the equation to eliminate the radical while keeping the equation balanced::

${\left(\sqrt{2 x + 3}\right)}^{2} = {6}^{2}$

$2 x + 3 = 36$

Next, subtract $\textcolor{red}{3}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$2 x + 3 - \textcolor{red}{3} = 36 - \textcolor{red}{3}$

$2 x + 0 = 33$

$2 x = 33$

Now, divide each side of the equation by $\textcolor{red}{2}$ to solve for $x$ while keeping the equation balanced:

$\frac{2 x}{\textcolor{red}{2}} = \frac{33}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = \frac{33}{2}$

$x = \frac{33}{2}$