# How do you solve 1+(x^2-5x-24)/(3x)=(x-6)/(3x) and check for extraneous solutions?

Dec 9, 2016

Put on a common denominator.

$\frac{3 x + {x}^{2} - 5 x - 24}{3 x} = \frac{x - 6}{3 x}$

${x}^{2} - 2 x - 24 = x - 6$

${x}^{2} - 3 x - 18 = 0$

$\left(x - 6\right) \left(x + 3\right) = 0$

$x = 6 \mathmr{and} - 3$

The restrictions of the initial equation are $x \ne 0$, so both solutions are valid.

Hopefully this helps!