# How do you solve sqrt(3x + 1) = -10?

Aug 24, 2015

$x = 33$

#### Explanation:

$\sqrt{3 x + 1} = - 10$

Square both sides of the equation.

(sqrt(3x+1))^2=(-10)^2

$3 x + 1 = 100$

Subtract $1$ from both sides.

$3 x = 100 - 1$$=$

$3 x = 99$

Divide both sides by $3$.

$x = \frac{99}{3}$$=$

$x = 33$

Check the answer by substituting $33$ for $x$.

$\sqrt{3 x + 1} = - 10$

$\sqrt{3 \cdot 33 + 1} = - 10$

$\sqrt{100} = - 10$

${\left(- 10\right)}^{2} = 100$

$\sqrt{100} = \pm 10$, so $33$ is a valid number for $x$.