Im having a little trouble with this one:

$\displaystyle

\frac{3}{x-2}-\frac{2}{x+3}=\frac{2}{5}

$

I can't find a decent "bottom half" fraction for this. Im thinking

$\displaystyle

5(x-2)(x+3)

$

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- Feb 3rd 2009, 09:58 AMPinguFractions and equations.
Im having a little trouble with this one:

$\displaystyle

\frac{3}{x-2}-\frac{2}{x+3}=\frac{2}{5}

$

I can't find a decent "bottom half" fraction for this. Im thinking

$\displaystyle

5(x-2)(x+3)

$ - Feb 3rd 2009, 10:17 AMBruce
Your denominator is good. Now set that as your denominator for each fraction. You will then have to multiply the top by the same thing you multiplied the bottom by. So:

$\displaystyle \frac {3 * 5(x+3)}{5(x - 2)(x + 3)} - \frac {2 * 5(x -2)}{5(x-2)(x+3)} = \frac {2 * (x-2)(x+3)}{5(x-2)(x+3)}$

then you simplify:

$\displaystyle \frac {15x + 45}{5(x-2)(x+3)} - \frac {10x - 20}{5(x-2)(x+3)} = \frac {2x^2 + 2x - 12}{5(x-2)(x+3)}$

Do the subtracting...

$\displaystyle \frac {5x + 65}{5(x-2)(x+3)} = \frac {2x^2 + 2x - 12}{5(x-2)(x+3)}$

Multiply both sides by the same denominator...

$\displaystyle 5x + 65 = 2x^2 + 2x - 12$

Can you do it from here? - Feb 3rd 2009, 10:20 AMPingu
Yes, I've tried that, and it turns out pretty messy. The correct number for X is 7, If my calculator is correct.

- Feb 3rd 2009, 10:22 AMtopher0805
- Feb 3rd 2009, 10:24 AMPingu
[quote=topher0805;

- Feb 3rd 2009, 10:31 AMhkerbest
$\displaystyle

-2x^2 + 3x = - 77

$

$\displaystyle

(-2x+3)(x-0) = - 77

$

X = 7 X = -5,5