# Thread: [SOLVED] Fractions: When to multiply the top and bottom, or only the top

1. ## [SOLVED] Fractions: When to multiply the top and bottom, or only the top

I have run into situations where I had to multiply either the top and bottom (numerator and denominator), or just the numerator.

So I was just wondering how I would know when I should use one or the other methods.

2. Originally Posted by fb280
I have run into situations where I had to multiply either the top and bottom (numerator and denominator), or just the numerator.

So I was just wondering how I would know when I should use one or the other methods.
$\frac{1}{2}+\frac{1}{3}=\left(\frac{3}{3}\right)\f rac{1}{2}+\left(\frac{2}{2}\right)\frac{1}{3}$

You multiply numerator and denominator when you want to change a fraction
to be compatible with another fraction, so you can combine them.

You are in fact multiplying by 1,
which does not change a fractions' value, just it's form.

$1=\frac{2}{2}=\frac{3}{3}=\frac{4}{4}=\frac{x}{x}= \frac{x+3}{x+3}=.......$

$\frac{x+1}{x}+\frac{x-1}{2x}=5$

In this case, you can combine the fractions as mentioned, but!

You may also multiply both sides by 2x to simplify
since the sides will still be equal and you no longer have fractions to contend with..

3. So basically what you're looking for when you multiply the numerators only on both sides, is when you know that it will cancel out all the denominators?

An example of what i'm stuck on would be:

$\frac{9}{y+2} - \frac{7}{y-2} = \frac{8}{y^2-4}$
that one multiplies only the top by (y-2)(y+2)

$\frac{1}{5} - \frac{1}{8} = \frac{1}{x}$
while this one multiplies both top and bottom to get 40 as a denominator

4. Originally Posted by fb280
So basically what you're looking for when you multiply the numerators only on both sides, is when you know that it will cancel out all the denominators?

An example of what i'm stuck on would be:

$\frac{9}{y+2} - \frac{7}{y-2} = \frac{8}{y^2-4}$
that one multiplies only the top by (y-2)(y+2)

$\frac{1}{5} - \frac{1}{8} = \frac{1}{x}$
while this one multiplies both top and bottom to get 40 as a denominator
Hi fb280,

you do have a choice of continuations,
here's how you can solve these in alternative ways...

$\frac{9}{y+2}-\frac{7}{y-2}=\frac{8}{y^2-4}$

Add the two fractions on the left, first obtaining a common denominator to make them compatible
(so that we may add or subtract the numerators)

$\frac{y-2}{y-2}\ \frac{9}{y+2}-\frac{y+2}{y+2}\ \frac{7}{y-2}=\frac{8}{y^2-4}$

Notice that $(y-2)(y+2)=y^2+2y-2y-4=y^2-4$

Hence the denominators on both sides are the same

$\frac{9(y-2)}{y^2-4}-\frac{7(y+2)}{y^2-4}=\frac{8}{y^2-4}$

$\frac{9(y-2)-7(y+2)}{y^2-4}=\frac{8}{y^2-4}$

If these are equal, since the denominators are the same, the numerators must also be equal.

$9(y-2)-7(y+2)=8$

Alternatively, we could have multiplied both sides of the fraction by (y-2)(y+2) as both sides will still be equal, since they were equal to begin with.
That will remove the denominators.

$\frac{(y+2)(y-2)9}{y+2}-\frac{(y+2)(y-2)7}{y-2}=\frac{8(y+2)(y-2)}{y^2-4}$

This gives $9(y-2)-7(y+2)=8$

which is the same result as before.

$\frac{1}{5}-\frac{1}{8}=\frac{1}{x}$

If you wish you may multiply both sides by 5, 8 and x to get

$\frac{40x}{5}-\frac{40x}{8}=\frac{40x}{x}$

$8x-5x=40\ \Rightarrow\ 3x=40\ \Rightarrow\ x=\frac{40}{3}$

Alternatively

$\frac{1}{5}-\frac{1}{8}=\frac{8}{8}\ \frac{1}{5}-\frac{5}{5}\ \frac{1}{8}=\frac{8}{40}-\frac{5}{40}=\frac{8-5}{40}=\frac{3}{40}$

$\frac{3}{40}=\frac{1}{x}\ \Rightarrow\ \frac{40}{3}=\frac{x}{1}=x$

5. Ok now I think it's starting to make more sense to me now. So both methods work in any given situations to get the denominators to be equal. I hope I got that right.