# Math Help - Addition with fractions

1. ## Re: Addition with fractions

[QUOTE=Plato;794073]Why do you complicate this so?

What a ridiculous thing to say... I made it complicated because I didn't know what I was doing, also your post doesn't help at all, you have not told me what you did the arrive at that answer. HallsOfIvy, topsquark and emakarov - Thank you for explaining.

2. ## Re: Addition with fractions

uperkurk...

$\frac{2}{5}$ is a fraction.

$\frac{3}{x(x+2)}$ is a fraction.

$\frac{2x^2 + 5x + 3}{x^2 - 10x + 25}$ is a fraction.

Don't let the complicated terms scare you. Essentially, they're just fractions. That means they have to follow all the rules of fractions. So whatever you'd normally do with $\frac{2}{3} + \frac{4}{5}$, you apply the same set of rules with $\frac{3x}{x-2} + \frac{4}{x+3}$.

3. ## Re: Addition with fractions

Originally Posted by uperkurk
Originally Posted by Plato
Why do you complicate this so?...
What a ridiculous thing to say... I made it complicated because I didn't know what I was doing, also your post doesn't help at all, you have not told me what you did the arrive at that answer. HallsOfIvy, topsquark and emakarov - Thank you for explaining.
I'm certain Plato meant no offense...he was simply demonstrating what is sometimes referred to as "cross multiplying."

If you are given an equation of the form:

$\frac{a}{b}=\frac{c}{d}$

Then, by this process, you may take as one side of the equation the product of the numerator on one side and the denominator on the other, and then then other side of the equation is the product of the other numerator-denominator pair. So, the equation above can then be written as:

$ad=bc$

4. ## Re: Addition with fractions

Fractional quantities are a by-product of wanting number systems "complete under division (except for 0)". As a result, multiplying with fractions is EASY, but adding them is HARD. The general rule is, of course:

a/b + c/d = (ad +bc)/(bd)

Sometimes, that is as good as it gets...it's not always possible to "factor out common factors" (although textbook problems often feature this, misleading students into an overly optimistic view of how things will turn out).

The equation:

a/b = c/d is often taken by definition to MEAN:

ad = bc...it's how we can tell that 1/2 is the same fraction as 2/4 (...because 1*4 = 2*2, see?). In other words "fractional expressions" aren't UNIQUE, we can always multiply top and bottom by the same thing (because a/a = 1, if a isn't 0), to get a "different-looking fraction". "Cancelling" is this process "in reverse" we are DIVIDING by a/a (and when you DIVIDE by b/c, you multiply by c/b (the reciprocal), and the reciprocal of a/a is, strangely enough, a/a again).

I've always felt that too much emphasis is placed upon putting things in "simplest form". It doesn't make an answer any more or less CORRECT, it just takes up less space on a piece of paper (and it might simplify calculations if you have to USE that answer later on). Fractions are often messy, and ugly-looking beasts. It may not seem so to you, dear uperkurk, but the problems you are tasked with solving have been tailored to shield you from the full blunt force of what fractional expressions can be.

5. ## Re: Addition with fractions

What you have to remember, uperkurk, is that the denominator has to be the same in all terms of the equation. It's no different from when you learned to add fractions in Arithmetic.

Example: $\displaystyle\frac{1}{4} + 1 +\displaystyle\frac{2}{3}$

$\displaystyle\frac{1}{4} = \displaystyle\frac{1(3)}{4(3)} = \displaystyle\frac{3}{12}$

$\displaystyle\frac{2}{3} = \displaystyle\frac{2(4)}{3(4)} = \displaystyle\frac{8}{12}$

$1= \displaystyle\frac{1(3)(4)}{1(3)(4)} = \displaystyle\frac{12}{12}$

$\displaystyle\frac{1}{4} + 1 + \displaystyle\frac{2}{3} = \displaystyle\frac{3}{12} +\displaystyle\frac{12}{12} + \displaystyle\frac{8}{12} = \displaystyle\frac{23}{12}$

You are doing the same thing when adding in algebraic equations. I believe this was explained in earlier posts, but I thought that one more example would help.

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