<s>[INST] How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, with two square root terms? So guys, my girlfriend is taking a college algebra class this summer and I figured I would help her study for her upcoming final because I am an engineering major and this kind of math would be easy for me. As we were doing problems, we came about one that I have no idea how to solve. It seems to be a straight forward "here's an equation, solve for $x$." Just one problem, I ended up not being able to solve for $x$, making me feel embarrassed since I was trying to help her and I am supposed to be the one who is "good at math." Anyway, it would be a grreat help if anyone can help me out on this. The equation is.... 
$$4\sqrt{x-3} - \sqrt{6x-17} = 3$$
The answer comes out to be $x = 7$ because I did it on my calculator. I first tried to just square each side to get rid of the square roots. But once I "FOILed" the left side, there were still square roots and things didn't look good for me. By the way, I am sorry I don't know how to format this equation correctly.
 [/INST] There is a trick for problems like this that I really love to do.
\begin{align}
   4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(multiply)} \\
   \text{Let } 4\sqrt{x-3} + \sqrt{6x-17} &= y \\
\hline
   16(x-3)-(6x-17) &= 3y \\
   16x-48 - 6x + 17 &= 3y \\
   y &= \dfrac{10x -31}{3} \\
\hline
   4\sqrt{x-3} + \sqrt{6x-17} &= \dfrac{10x -31}{3} &\text{(so now we have this)} \\
   4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(subtract them)}\\
\hline
   2\sqrt{6x-17} &= \dfrac{10x -40}{3} \\
   \sqrt{6x-17} &= \dfrac{5x - 20}{3} \\
   3\sqrt{6x-17} &= 5x - 20 \\
   54x - 153 &= 25x^2 - 200x + 400 \\
   25x^2 - 254x + 553 &= 0 \\
   (x-7)(25x - 79) &= 0 \\
   x &\in \left\{ 7, \dfrac{79}{25}\right\}
\end{align}
Check for extraneous roots.
When $x = 7$
$$ (4\sqrt{x-3} - \sqrt{6x-17})= 4\cdot 2 - 5 = 3$$
When $x = \dfrac{79}{25}$
$$ (4\sqrt{x-3} - \sqrt{6x-17})= 4\cdot \dfrac 25 - \dfrac 75 
= \color{red}{\frac 15}$$
So we reject $x = \dfrac{79}{25}$ as an extraneous root and accept $x=7$,
</s>