# Express sum as single algebraic fraction

1. May 4, 2007

### thomas49th

1. The problem statement, all variables and given/known data

Express:

$$\frac{1}{x-2}+\frac{2}{x+4}$$

3. The attempt at a solution

Well I got x² + 2x - 11 = 0

but I think that is wrong

2. May 4, 2007

### Integral

Staff Emeritus
How did you get your solution? Show us!

3. May 4, 2007

### thomas49th

1 + 2 = (x-2)(x+4)
3 = x² + 2x - 8
then take 3 from both sides gives you the answer I previsouly posted... bust that doesn't seem right as how do I kow

[ex]\frac{1}{x-2}+\frac{2}{x+4}[/tex] = 1

is equal to 1

4. May 4, 2007

### Integral

Staff Emeritus
To clear fractions you must cross multiply.

$$\frac a b + \frac c d = \frac {ad + bc} {bd}$$

is that how you did it?

5. May 4, 2007

### Feldoh

It shouldn't equal anything :-/

6. May 4, 2007

### VietDao29

Nope, the first line is just sooooo wrong, you cannot do that.
What you should do is to make common denominator, or in other words, cross multiply, as Integral has pointed out:
$$\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad + bc}{bd}$$
Ok, I'll give you an example:
$$\frac{1}{x - 5} + \frac{2}{x} = \frac{x + 2 (x - 5)}{x (x - 5)} = \frac{3x - 10}{x ^ 2 - 5x}$$.
Can you get it? :)

7. May 5, 2007

### thomas49th

$$\frac{3x}{x^{2}+2x-8}$$

is what I got

8. May 5, 2007

### Hootenanny

Staff Emeritus
You might want to try that again. Here, I'll start for you;

$$\frac{1}{x-2}+\frac{2}{x+4} = \frac{1(x+4)+2(x-2)}{(x-2)(x+4)}$$

Can you simplify that any?

9. May 5, 2007

### thomas49th

$$\frac{x+4+2x-4}{x^{2}+2x-8}$$

which is hen simplified to

$$\frac{3x}{x^{2}+2x-8}$$

10. May 5, 2007

### Hootenanny

Staff Emeritus
Sorry, my bad. I had a sign error, you are of course correct!