Question

How do you solve `5/(x-3)=(2x)/(x^2-9)`?

Answer 1

`x = -5`

Explanation:

Note that `x` must be different from `\pm 3`. In fact, the first denominator vanished is `x=3`, while the second vanishes if `x=\pm 3`.

Note that `x^2-9=(x+3)(x-3)`, so if we multiply and divide the left hand side by `x+3` (and this is legit since `x \ne -3`) we get

`\frac{5}{x-3}*\frac{x+3}{x+3} = \frac{2x}{x^2-9}`

The left hand side can be evaluated as we noted above, so we have

`\frac{5(x+3)}{x^2-9} = \frac{2x}{x^2-9}`

We can multiply both sides by the denominator, since it's the same for both and it can't be zero:

`5(x+3) = 2x \iff 5x+15 = 2x \iff 3x = -15 \iff x = -5`