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)
$
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)
$
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?