How do you solve $(x)/(x-2) – (8)/(x+3) = 10/(x^2+x-6)$ ?

Question

3130 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T14:29:01

$x = 3, x!= -3, 2$

Start by factoring everything to see what the denominators are and what we can cancel out.

$=>x/(x - 2) - 8/(x + 3) = 10/((x + 3)(x - 2))$

The common denominator will be $(x + 3)(x - 2)$ .

$=>(x(x + 3))/((x + 3)(x - 2)) - (8(x - 2))/((x + 3)(x - 2)) = 10/((x + 3)(x - 2))$

We can now cancel the denominators and solve as a regular quadratic.

$=>x^2 + 3x - 8x + 16 = 10$

$=>x^2 - 5x + 6 = 0$

$=>(x - 3)(x - 2) = 0$

$=>x = 3 and 2$

However, the $x = 2$ is extraneous, since it is a restriction on the variable (it makes the denominator equal to $0$ and therefore undefined).

Checking in the original equation, $x = 3$ works.

Hopefully this helps!

How do you solve (x)/(x-2) – (8)/(x+3) = 10/(x^2+x-6)(x)/(x-2) – (8)/(x+3) = 10/(x^2+x-6)?