How do you solve (3x)/(x-4)=(2x)/(x-3)+6/(x^2-7x+12) and check for extraneous solutions?

Aug 25, 2016

$x = - 2$

Explanation:

Start by factoring the denominator.

$\frac{3 x}{x - 4} = \frac{2 x}{x - 3} + \frac{6}{\left(x - 4\right) \left(x - 3\right)}$

Put on a common denominator:

$\frac{3 x \left(x - 3\right)}{\left(x - 4\right) \left(x - 3\right)} = \frac{2 x \left(x - 4\right)}{\left(x - 3\right) \left(x - 4\right)} + \frac{6}{\left(x - 3\right) \left(x - 4\right)}$

$3 {x}^{2} - 9 x = 2 {x}^{2} - 8 x + 6$

${x}^{2} - x - 6 = 0$

$\left(x - 3\right) \left(x + 2\right) = 0$

$x = 3 \mathmr{and} - 2$

Let's finally note restrictions on the variable. This is done by setting the denominators to $0$ and solving for $x$.

$x - 3 = 0 \to x = 3$

AND

$x - 4 = 0 \to x = 4$

Hence, $x \ne 3 \mathmr{and} x \ne 4$

Since $x = 3$ contradicts these restrictions, $x = 3$ is an extraneous solution. The only actual solution is $x = - 2$. Our solution set is therefore $\left\{- 2\right\}$.

Hopefully this helps!