# Simple math: What am I doing wrong?

1. Dec 8, 2007

### Holocene

Consider the addition of 2 fractions:

$$\displaystyle{\frac{7}{x - 3} + \frac{1}{3 - x}}$$

We know we must find a least common denominator in order to add the fractions, so let's consider them:

We have x - 3, and we also have 3 - x.

Suppose we rewrote the first denominator as -(3 - x). In other words: x - 3 = -(3 - x)

Our denominators would then be -(3 - x) & (3 - x). Note we haven't changed the value of the denominators, we have simply rewritten them.

The fraction problem would then become:

$$\displaystyle{\frac{7}{-(3 - x)} + \frac{1}{3 - x}}$$

Note that when a minus sign is placed in a fraction, it can be placed in either the numberator or the denominator, as it does not matter. So, we can have identical denominators for the purpose of addition by simply moving the minus sign to the numberator in the first fraction:

$$\displaystyle{\frac{-7}{3 - x} + \frac{1}{3 - x} = \frac{-6}{3 - x}}$$

My question is, why is this wrong?

The book instead rewrites the OTHER denominator, and the final problem looks like this:

$$\displaystyle{\frac{7}{x - 3} + \frac{1}{3 - x} = \frac{7}{x - 3} + \frac{1}{-(x - 3)} = \frac{7}{x - 3} + \frac{-1}{x - 3} = \frac{6}{x - 3}}$$

What am I doing wrong?

Last edited: Dec 8, 2007
2. Dec 8, 2007

You're answer is the same as the books, but you have written the minus sign on the top, and they have written it on the bottom.

3. Dec 8, 2007

### Holocene

Damn I just realized that...:shy:

Amazing how it helps to look at the whole thing on the screen like that.

Thanks!

4. Dec 8, 2007

### Holocene

Also, my answer had a negative term in the fraction, and theirs did not, so I guess theirs is more of the "correct" way, even if the values are in fact the same?

5. Dec 8, 2007

### BSMSMSTMSPHD

It's not "more correct" (not even really sure what that means!) but I personally find it more concise to write answers with the minimum number of symbols, including negative signs. Your answer is fine.

6. Dec 8, 2007

### arildno

Holocene:

Let us prove that:
$$\frac{-a}{b}=\frac{a}{-b}=-\frac{a}{b}$$

First off:
$$\frac{a}{b}+\frac{-a}{b}=a*\frac{1}{b}+(-a)*\frac{1}{b}=(a+(-a))*\frac{1}{b}=0*\frac{1}{b}=0$$
Hence, we have: $$\frac{-a}{b}=-\frac{a}{b}$$, since by

1.definiton of -a/b, we have:
$$\frac{a}{b}+(-\frac{a}{b})=0$$
and
2. We may prove that additive inverses are unique.

Furthermore:
$$\frac{a}{-b}=\frac{1*a}{(-1)*b}=\frac{1}{(-1)}*\frac{a}{b}=(-1)*\frac{a}{b}=-\frac{a}{b}$$
by employing a few derivable results.