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Problem: $\sqrt{x+3} -\sqrt{x-2} = \sqrt{6x-11}$ So I squared both sides: $(\sqrt{x+3} -\sqrt{x-2})^2 = (\sqrt{6x-11})^2$ I was left with: $x+3 - x-2 = 6x-11$

However, this is incorrect since I must first simplify within the parenthesis, I am having trouble with advanced precalculus problems because I often make algebraic mistakes such as that above and I would like to ask for any advice. Does anyone have a good source where I can review all the algebriac rules? I feel embarrassed asking this but I am seriously being held back by this.

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    It's alright. Everybody makes mistakes. After all, you learn from your own mistakes only. But don't give up and keep practicing. Also, just make sure, at each step, question yourself, if it's valid or not? if it assumes something implicitly or not? and so on, to make less mistakes. – 2017-02-11
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    after squaring $x+3 + x- 2 - 2 \sqrt {(x+3)(x-2)} = 6x -11 $ – 2017-02-11

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I join the other answers and suggest to practice. As another good way to understand algebraic manipulations, you should see the geometric side of things.

Some (if not most) school programs separate 'algebra-precalculus' from geometry for no reason whatsoever. It's even worse when trigonometry is also presented like a separate subject. They all describe the same objects, just from different sides!

Like in your case. Initially you have the problem with three line segments:

enter image description here

You can see that one of them is a sum of two others.

$$\sqrt{x-2}+\sqrt{6x-11}=\sqrt{x+3}$$

To get rid of square roots you square everything like this:

enter image description here

You can see that you now have two additional rectangles of the same area $S_4$ and $S_5$ which still have square roots.

$$x-2+2\sqrt{x-2}\sqrt{6x-11}+6x-11=x+3$$

What you should do now, is to get everything but the square root on one side, and the square root part on the other side:

$$2\sqrt{x-2}\sqrt{6x-11}=16-6x$$

In this case you can divide both sides by $2$:

$$\sqrt{x-2}\sqrt{6x-11}=8-3x$$

And now square again:

$$(x-2)(6x-11)=(8-3x)^2$$

You get a quadratic equation, which again can be better understood using geometry.

Beware: the quadratic equation has two solutions, only one of them fits the original equation with square roots. You should check which one by replacing $x$ with it and seeing if the equality is true in each case.

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Practice. You just have to learn and practice. Your problem here is that $(a+b)^2 \neq a^2 + b^2$, which you seem to have assumed. This is a common mistake and is nothing to beat yourself up about.

As far as a single reference on this? I recommend taking a class or reading a textbook, though AlgebraRules seems to be a nice source

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    Thanks, unfortunately I must now go back and study algebra rules before returning to calc :( – 2017-02-11