# How do you solve for x: 2/(x-3) - 3/(x+3) = 12/((x^2)-9) ?

Feb 25, 2017

$\frac{2}{x - 3} - \frac{3}{x + 3} = \frac{12}{{x}^{2} - 9}$

$\implies \frac{2 \left(x + 3\right) - 3 \left(x - 3\right)}{\left(x + 3\right) \left(x - 3\right)} = \frac{12}{{x}^{2} - 9}$

$\implies \frac{2 x + 6 - 3 x + 9}{\cancel{{x}^{2} - 9}} = \frac{12}{\cancel{{x}^{2} - 9}}$

$\implies - x + 15 = 12$

$\implies x = 3$

Verification: Put $x = 3$

$L . H . S = \frac{2}{x - 3} - \frac{3}{x + 3} = \frac{2}{3 - 3} - \frac{3}{3 + 3} = \frac{2}{0} - \frac{3}{6}$

Here we can see $\frac{2}{0}$, in Mathematics, division by 0 is not allowed.
Hence, the root $x = 3$ is not a solution to the given equation.
Hence, no solution exists for the given equation.