Question

How do you solve `(3x-3)/(x^2-13x+40)-(x-6)/(x^2-22x+112)=(x-9)/(x^2-19x+70)` and check for extraneous solutions?

Answer 1

`x in {20, -3}`

Explanation:

Factor the denominators.

`x^2 - 13x + 40 = (x - 8)(x- 5)`
`x^2 - 22x + 112 = (x - 8)(x - 14)`
`x^2 - 19x + 70 = (x - 14)(x - 5)`

Now, put each fraction on a common denominator.

`((3x- 3)(x - 14))/((x- 8)(x - 5)(x - 14)) - ((x- 6)(x - 5))/((x - 8)(x - 5)(x - 14)) = ((x- 9)(x - 8))/((x- 8)(x - 5)(x - 14))`

`3x^2 - 3x - 42x + 42 - (x^2 - 11x + 30) = x^2 - 17x + 72`

`3x^2 - 45x + 42 - x^2 + 11x - 30 = x^2 - 17x + 72`

`x^2 - 17x - 60 = 0`

`(x- 20)(x + 3) = 0`

`x = 20 or -3`

Our initial restrictions were `x != 5, 8, 14`, so all of our solutions are acceptable.

Hopefully this helps!