# How do you solve 2+sqrt(5x-1)>5?

May 7, 2018

color(blue)((2,oo)

#### Explanation:

$2 + \sqrt{5 x - 1} > 5$

Subtract 2 from both sides of the inequality sign:

$2 - \textcolor{red}{2} + \sqrt{5 x - 1} > 5 - \textcolor{red}{2}$

$\sqrt{5 x - 1} > 3$

Square both sides:

${\left(\sqrt{5 x - 1}\right)}^{2} > {3}^{2}$

$5 x - 1 > 9$

$5 x - 1 + \textcolor{red}{1} > 9 + \textcolor{red}{1}$

$5 x > 10$

Divide both sides by 5:

$\frac{5 x}{5} > \frac{10}{5}$

Cancelling:

$\frac{\cancel{5} x}{\cancel{5}} > \frac{\cancel{10} 2}{\cancel{5}}$

$x > 2$

We can write the range of solutions in interval notation as:

$\left(2 , \infty\right)$