Question

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

Answer 1

`x = 24 " or " x = -1`

Explanation:

Identify the restrictions on `x` before you even solve the equation.
The denominators may not be equal to 0.

`3x +5 != 0" and "x-3!=0`
`3x !=-5color(white)(xxxxxxxxxxx)x!=3`
`x != -5/3`

As there is one fraction on each side of the equal sign you can cross -multuiply.

`(x-2)(x-3)= 6(3x+5)`

`x^2 -3x-2x+6 = 18x+30" "larr` a quadratic, so set =0

`x^2 -5x-18x+6-30= 0`

`x^2 -23x-24=0" "larr` find factors of 24 which differ by 23

`(x-24)(x+1)=0`

`x = 24 " or " x = -1`

Both solutions are valid.