Solve for x, where x is a real number.
[tex] \sqrt{6x+2} =\sqrt{9x-8} [/tex]
If there is more than one solution, separate them with commas.
If there is no solution, click on "No solution"
Please explain how you got your answer!

Respuesta :

Hello!

First of all, let's undo the square roots by squaring both sides.

6x+2=9x-8

We subtract 9x from both sides.

-3x+2=-8

We subtract 2 from both sides.

-3x= -10

We divide both sides by -3.

x=10/3

I hope this helps!

[tex] \sqrt{6x+2}=\sqrt{9x-8}\qquad \left(x\geq-\dfrac{1}{3} \wedge x\geq \dfrac{8}{9}\implies x\geq \dfrac{8}{9}\right)\\\\
6x+2=9x-8\\\\
3x=10\\\\
x=\dfrac{10}{3} [/tex]