Question

Solve the equation.

Answer 1

Invisible math?

Answer 2

Try adding 2 to both sides, and then squaring both sides

Answer 3

Try writing \(6 - 3x \ge 0 \implies x \le 2\)

Answer 4

Example problem: \(\color{#000000}{\displaystyle\sqrt{2x-6}-7=1 }\)

Add 7 to both sides to isolate the root.

\(\color{#000000}{\displaystyle\sqrt{2x-6}-7\color{red}{+7}=1\color{red}{+7} }\)

Note that 7's on the left side cancel.

\(\color{#000000}{\displaystyle\sqrt{2x-6}=8 }\)

Raise both sides to the second power.

\(\color{#000000}{\displaystyle\color{red}{(\color{black}{\sqrt{2x-6}})^2}=\color{red}{(\color{black}{8})^2}}\)

you know that \((\sqrt{~a~})^2=a\), so same here.
The square root goes off the stage:) (And 8²=64)
So, you get,

\(\color{#000000}{\displaystyle 2x-6=64 }\)

Add 6 to both sides to isolate the "2x"

\(\color{#000000}{\displaystyle 2x-6\color{red}{+6}=64\color{red}{+6} }\)

Notice, just as we cancel 7's at the beginning, we cancel the 6's.

\(\color{#000000}{\displaystyle 2x=70}\)

We have isolated "2x", now we can solve for x, by dividing by 2 on both sides.

\(\color{#000000}{\displaystyle 2x\color{red}{\div 2}=70\color{red}{\div 2}}\)

and you get (as the answer)

\(\color{#000000}{\displaystyle x=35}\)

Answer 5

(there's the example... without mistakes)