# How do you solve sqrt(7r+2)+3=7 and check your solution?

Aug 9, 2017

See a solution process below:

#### Explanation:

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

$\sqrt{7 r + 2} + 3 - \textcolor{red}{3} = 7 - \textcolor{red}{3}$

$\sqrt{7 r + 2} + 0 = 4$

$\sqrt{7 r + 2} = 4$

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

${\left(\sqrt{7 r + 2}\right)}^{2} = {4}^{2}$

$7 r + 2 = 16$

Then, subtract $\textcolor{red}{2}$ from each side of the equation to isolate the $r$ term while keeping the equation balanced:

$7 r + 2 - \textcolor{red}{2} = 16 - \textcolor{red}{2}$

$7 r + 0 = 14$

$7 r = 14$

Now, divide each side of the equation by $\textcolor{red}{7}$ to solve for $r$ while keeping the equation balanced:

$\frac{7 r}{\textcolor{red}{7}} = \frac{14}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} r}{\cancel{\textcolor{red}{7}}} = 2$

$r = 2$

To validate the solution substitute $\textcolor{red}{2}$ for $\textcolor{red}{r}$ in the original equation and calculate the result to ensure both sides of the equation are equal (remember, the square root of a number produces a positive and negative result):

$\pm \sqrt{7 \textcolor{red}{r} + 2} + 3 = 7$ becomes:

$\pm \sqrt{\left(7 \cdot \textcolor{red}{2}\right) + 2} + 3 = 7$

$\pm \sqrt{14 + 2} + 3 = 7$

$\pm \sqrt{16} + 3 = 7$

$- 4 + 3 = 7$ and $4 + 3 = 7$

$- 1 \ne 7$ and $7 = 7$

The negative result of the radical is an extraneous solution.

The positive result shows the solution is correct.