Question

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

Answer 1

`x = -2`

Explanation:

Start by factoring the denominator.

`(3x)/(x - 4) = (2x)/(x - 3) + 6/((x - 4)(x - 3))`

Put on a common denominator:

`(3x(x - 3))/((x - 4)(x - 3)) = (2x(x - 4))/((x - 3)(x - 4)) + 6/((x - 3)(x - 4))`

`3x^2 - 9x = 2x^2 - 8x + 6`

`x^2 - x - 6 = 0`

`(x - 3)(x + 2) = 0`

`x = 3 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 -> x = 3`

AND

`x - 4 = 0 -> x = 4`

Hence, `x!=3 and x!= 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 `{-2}`.

Hopefully this helps!