# How do you solve 2sqrt(3x+5)+7=16?

May 9, 2017

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{7}$ from each side of the equation to isolate the radical term while keeping the equation balanced:

$2 \sqrt{3 x + 5} + 7 - \textcolor{red}{7} = 16 - \textcolor{red}{7}$

$2 \sqrt{3 x + 5} + 0 = 9$

$2 \sqrt{3 x + 5} = 9$

Next, divide each side of the equation by $\textcolor{red}{2}$ to isolate the radical while keeping the equation balanced:

$\frac{2 \sqrt{3 x + 5}}{\textcolor{red}{2}} = \frac{9}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt{3 x + 5}}{\cancel{\textcolor{red}{2}}} = \frac{9}{2}$

$\sqrt{3 x + 5} = \frac{9}{2}$

Then, square each side of the equation to eliminate the radical while keeping the equation balanced:

${\left(\sqrt{3 x + 5}\right)}^{2} = {\left(\frac{9}{2}\right)}^{2}$

$3 x + 5 = \frac{81}{4}$

Next, subtract $\textcolor{red}{5}$ from each side of the equation to isolate the $x$ term:

$3 x + 5 - \textcolor{red}{5} = \frac{81}{4} - \textcolor{red}{5}$

$3 x + 0 = \frac{81}{4} - \left(\frac{4}{4} \times \textcolor{red}{5}\right)$

$3 x = \frac{81}{4} - \frac{20}{4}$

$3 x = \frac{61}{4}$

$n o w , \div i \mathrm{de} e a c h s i \mathrm{de} o f t h e e q u a t i o n b y$color(red)(3)$\to s o l v e f \mathmr{and}$x# while keeping the equation balanced:

$\frac{3 x}{\textcolor{red}{3}} = \frac{\frac{61}{4}}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}} = \frac{61}{12}$

$x = \frac{61}{12}$