Express sum as single algebraic fraction

[thomas49th]

Homework Statement

Express:

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

The Attempt at a Solution

Well I got x² + 2x - 11 = 0

but I think that is wrong

 

[Integral]

How did you get your solution? Show us!

 

[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

 

[Integral]

To clear fractions you must cross multiply.

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

is that how you did it?

 

[Feldoh]

Homework Statement

Express:

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

The Attempt at a Solution

Well I got x² + 2x - 11 = 0

but I think that is wrong

It shouldn't equal anything :-/

 

[VietDao29]

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

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? :)

 

[thomas49th]

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

is what I got

 

[Hootenanny]

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?

 

[thomas49th]

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

which is hen simplified to

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

 

[Hootenanny]

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