## Elementary Algebra

We first add 3 to both sides of the equation to obtain that $\sqrt{6x-5}=3$. Recall, in order to cancel out a square root, we square both sides of the equation. Thus, we square both sides of the equation to obtain that 6x-5=9. We add 5 on both sides and then divide both sides of the equation by 6 to isolate x and obtain: x=14/6=7/3. We now must check our solution: $\sqrt{6(7/3)-5}-3 = 0 \\\\\ \sqrt {9}-3=0 \\\\ 0=0$ This is true, so x=7/3.