Question

How to solve `36/[x^2-9]= (2x)/[x + 3]- 1/[1]`?

Answer 1

`x = 9`

The extraneous solution `x =-3` is rejected.

Explanation:

`36/[x^2-9]= (2x)/[x + 3]- 1/[1]`

`1/1` can be anything, make the right side denominators equal.

`36/[color(red)(x^2-9)]= color(blue)((2x)/[x + 3]- (x+3)/(x+3))" "larr` factorise and simplfy

`36/color(red)[(x+3)(x-3)]=color(blue)( (2x -(x+3))/(x+3))`

`36/[(x+3)(x-3)]= (2x -x-3)/(x+3)`

`36/[(x+3)(x-3)]= (x-3)/(x+3)" "color(forestgreen)(xx (x+3))`

`(36color(forestgreen)(xx cancel((x+3))))/[cancel((x+3))(x-3)]= ((x-3)color(forestgreen)(xxcancel ((x+3))))/cancel((x+3))" "larr` cancel

`36/((x-3)) = (x-3)" "larr` cross multiply

`(x-3)^2 =36" "larr` square root both sides

`x- 3= +-6`

If `x-3 =+ 6" "rarr x = 9`

If `x-3=-6" "rarr x = -3`

However, `x=-3` will make the denominators 0, so this is an extraneous solution which we will reject.