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Algebra Problems Go-Through 3

Author: oLahav

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Algebra Problems Go-Through:Part 3, annoying ones

As a follow-up to the popular Welcome to Algebra series, here comes a whole new series of lessons, going through algebra sample problems step-by-step for you to follow. We now step away from word problems and look at annoying expressions and simplification.

Say you've got one nasty expression. What do you do?

To deal with complicated algebraic expressions, you have to master the concept of cleaning things up. In math, this is called simplification. Start off by remembering that you can collect like terms and factor. For example: 5xy+10xw+7y+14w=5x(y+2w)+7(y+2w)=(5x+7)(y+2w) . You can make sure this is true by plugging in values for your variables and checking that both sides give you the same answer.

A really useful application you've already seen is quadratic equations. For example, consider this:
10x ^ 2+27x- 28=10x ^ 2+35x- 8x- 28 =5x(2x+7) -4(2x+7)=(5x-4)(2x+7) .

And don't forget the nice formulas:
a ^ 2+2ab+b ^ 2=(a+b) ^ 2
a ^ 2- 2ab+b ^ 2=(a-b) ^ 2
a ^ 2- b ^ 2=(a+b)(a-b) .

These types of simplification takes mostly practice to master, and there's lots of resources on the web, like here.

And what if it's uglier? Like fractions, everybody hates fractions

Something to watch for are those questions with expressions that look like fractions, called rational expressions. These aren't too bad as long as you remember rule #1: you can cancel out factors, but not individual terms. Let's simplify this: $\frac{2x ^ 2+13x+20}{2x ^ 2+17x+20}$ . You may want to cancel out the 2x ^ 2 at front there, because both the up and down have it, but you can't do that- it's not a factor, it's just an individual part of the expression. So how do we do it? Factor each side, and then cancel out the factors:
$\frac{2x ^ 2+13x+20}{2x ^ 2+17x+20}=\frac{(2x+5)(x+4)}{(2x+5)(x+6)}=\frac{x+4}{x+6}$ .

See, now it's a lot nicer. And it works, plugging any value of x will give you the same answer in both versions, unless $x= \frac{-2}{5}$ , since in that case the expression isn't defined. You have to remember, when giving your final answer, to state these things, called limits on domain.

How about square roots? Nobody likes those either

Roots are not fun. Root laws work like exponent laws, which you should be aware of. But a mathematical convention requires you to never leave roots in the denominator of an expression, so how do get rid of them? Start off easy, with say $\frac{1}{\sqrt{x}}$ . An instinct can tell you to multiply this by itself, but this changes the expression. To simplify, multiply by 1. This doesn't do anything, but if you write $1=\frac{\sqrt{x}}{\sqrt{x}}$ , you can see how it helps. Multiply top and bottom to get $\frac{\sqrt{X}}{x}$ , which may not look nicer but it mathematically more appropriate.

The same principle works with more complicated expressions. Take a look: $\frac{2x}{\sqrt{x- 3}}=\frac{2x* \sqrt{x- 3}}{\sqrt{x- 3}* \sqrt{x- 3}}= \frac{2x\sqrt{x- 3}}{x- 3}$ . Not so bad, and now you're following conventions in the right way.