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

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Answer 1 · 2017-03-24T17:07:35

$x = 9$

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

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

$1/1$ 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.

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