[SOLVED] Help with solve equation

**onymousillusion** · March 16th 2010, 05:09 PM

$2x/x^2-4 = 4/x^2-4 - 3/x+2$

The back of the book said that it's no solution but I get an answer of x = -2/5
I plug it back it make sure it doesn't make the denominator go to 0 and it doesn't so I'm not doing something right.... any help would be appreciated

**craig** · March 16th 2010, 05:29 PM

Originally Posted by onymousillusion

$2x/x^2-4 = 4/x^2-4 - 3/x+2$

The back of the book said that it's no solution but I get an answer of x = -2/5
I plug it back it make sure it doesn't make the denominator go to 0 and it doesn't so I'm not doing something right.... any help would be appreciated

Could you write out your equation using the fraction latex tag, it's unclear what exactly is a fraction and what isn't.

For example, using \frac{a}{b} to get $\frac{a}{b}$

**onymousillusion** · March 16th 2010, 05:40 PM

Thanks for the tip didn't no you could do that...
well here is all the step i did.
$ \frac{2x}{x^2-4} = \frac{4}{x^2-4} - \frac{3}{x+2} $

$ \frac{2x}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} - \frac{3}{x+2} $

$ \frac{2x(x-2)(x+2)}{(x-2)(x+2)} = \frac{4(x-2)(x+2)}{(x-2)(x+2)} = \frac{3(x+2)(x-2)}{x+2} $

than I canceled and got

$ 2x = 4 - 3x - 6 $

and you can see how I got x = -2/5

**harish21** · March 16th 2010, 05:46 PM

Originally Posted by onymousillusion

$2x/x^2-4 = 4/x^2-4 - 3/x+2$

The back of the book said that it's no solution but I get an answer of x = -2/5
I plug it back it make sure it doesn't make the denominator go to 0 and it doesn't so I'm not doing something right.... any help would be appreciated

I assume your question is

$\frac{2x}{x^{2}-4} = \frac{4}{x^{2}-4} - \frac{3}{x+2}$

or, $\frac{2x}{x^{2}-4} - \frac{4}{x^{2}-4} = - \frac{3}{x+2}$

or, $\frac{2x-4}{x^{2}-4} = - \frac{3}{x+2}$

or, $\frac{2(x-2)}{(x+2)(x-2)} = - \frac{3}{x+2}$

or, $\frac{2}{x+2} = - \frac{3}{x+2}$

or, $2 = - \frac{3(x+2)}{x+2}$

or, $2 = -3$

which is why there is no solution

**craig** · March 16th 2010, 05:50 PM

Originally Posted by onymousillusion

Thanks for the tip didn't no you could do that...
well here is all the step i did.
$ \frac{2x}{x^2-4} = \frac{4}{x^2-4} - \frac{3}{x+2} $

$ \frac{2x}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} - \frac{3}{x+2} $

$ \frac{2x(x-2)(x+2)}{(x-2)(x+2)} = \frac{4(x-2)(x+2)}{(x-2)(x+2)} = \frac{3(x+2)(x-2)}{x+2} $

than I canceled and got

$ 2x = 4 - 3x - 6 $

and you can see how I got x = -2/5

From what I can see your equation should be $2x = 4 - 3x -(-6)$ , which will lead you to $x=2$

**onymousillusion** · March 16th 2010, 05:51 PM

I see what I did wrong thanks again everyone....

**harish21** · March 16th 2010, 05:58 PM

Originally Posted by craig

From what I can see your equation should be $2x = 4 - 3x -(-6)$ , which will lead you to $x=2$

I think the answers x=-2/5 and x=2 are both incorrect!

Does x=2 satisfy the given equation? you get 0 = -3/4

there is no solution to this equation

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