# How do you solve sqrt(6x-5)+10=3?

Mar 12, 2017

See the entire solution process below:

#### Explanation:

First, subtract $\textcolor{red}{10}$ from each side of the equation to isolate the square root term while keeping the equation balanced:

$\sqrt{6 x - 5} + 10 - \textcolor{red}{10} = 3 - \textcolor{red}{10}$

$\sqrt{6 x - 5} + 0 = - 7$

$\sqrt{6 x - 5} = - 7$

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

${\left(\sqrt{6 x - 5}\right)}^{2} = {\left(- 7\right)}^{2}$

$6 x - 5 = 49$

Then, add $\textcolor{red}{5}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$6 x - 5 + \textcolor{red}{5} = 49 + \textcolor{red}{5}$

$6 x - 0 = 54$

$6 x = 54$

Now, divide each side of the equation by $\textcolor{red}{6}$ to solve for $x$ while keeping the equation balanced:

$\frac{6 x}{\textcolor{red}{6}} = \frac{54}{\textcolor{red}{6}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} x}{\cancel{\textcolor{red}{6}}} = 9$

$x = 9$

Mar 12, 2017

no solution.

#### Explanation:

$\textcolor{b l u e}{\text{Isolate the square root}}$ by subtracting 10 from both sides.

$\sqrt{6 x - 5} \cancel{+ 10} \cancel{- 10} = 3 - 10$

$\Rightarrow \sqrt{6 x - 5} = - 7$

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

${\left(\sqrt{6 x - 5}\right)}^{2} = {\left(- 7\right)}^{2}$

$\Rightarrow 6 x - 5 = 49$

$6 x \cancel{- 5} \cancel{+ 5} = 49 + 5$

$\Rightarrow 6 x = 54$

divide both sides by 6

$\frac{\cancel{6} x}{\cancel{6}} = \frac{54}{6}$

$\Rightarrow x = 9$

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

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

$\text{left side } = \sqrt{\left(6 \times 9\right) - 5} + 10$

$= \textcolor{w h i t e}{\le f t s i \mathrm{de}} = \sqrt{49} + 10$

=color(white)("left side)=7+10

$\textcolor{w h i t e}{\times \times \times \times} = 17 \ne 3$

$\Rightarrow \text{ there is no solution}$