# How do you solve sqrt(6-b)=sqrt(b+10) and check your solution?

Jul 25, 2017

See a solution process below:

#### Explanation:

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

${\left(\sqrt{6 - b}\right)}^{2} = {\left(\sqrt{b + 10}\right)}^{2}$

$6 - b = b + 10$

Next, add $\textcolor{red}{b}$ and subtract $\textcolor{b l u e}{10}$ from each side of the equation to isolate the $b$ term while keeping the equation balanced:

$6 - b + \textcolor{red}{b} - \textcolor{b l u e}{10} = b + 10 + \textcolor{red}{b} - \textcolor{b l u e}{10}$

$6 - \textcolor{b l u e}{10} - b + \textcolor{red}{b} = b + \textcolor{red}{b} + 10 - \textcolor{b l u e}{10}$

$- 4 - 0 = 1 b + \textcolor{red}{1 b} + 0$

$- 4 = \left(1 + \textcolor{red}{1}\right) b$

$- 4 = 2 b$

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

$- \frac{4}{\textcolor{red}{2}} = \frac{2 b}{\textcolor{red}{2}}$

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

$- 2 = b$

$b = - 2$

To validate the answer we will substitute $\textcolor{red}{- 2}$ for each occurrence of $\textcolor{red}{b}$ in the original equation and determine if both sides of the equation are equal:

$\sqrt{6 - \textcolor{red}{b}} = \sqrt{\textcolor{red}{b} + 10}$ becomes:

$\sqrt{6 - \textcolor{red}{- 2}} = \sqrt{\textcolor{red}{- 2} + 10}$

$\sqrt{6 + \textcolor{red}{2}} = \sqrt{\textcolor{red}{- 2} + 10}$

$\sqrt{8} = \sqrt{8}$

Both sides of the equation are equal so the answer is correct.