How do you subtract #9/(x^2-9) - 5/ (x+3)#?

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Answer 1 · 2015-10-29T11:55:20

I have shown every step in 'over the top' detail so you can see where everything comes from.

#- (5x-24)/(x^2-9)#

When presented with questions you look for links which means you have to build up a 'toolbox' of remembered facts and techniques:

Given:# 9/(x^2-9) - 5/(x+3)# .............(1)

Consider #x^2-9# ....................(2)

But #9 = 3^2# .....................(3)

Substitute (3) into (2) giving:

#x^2 - 3^3#

But #(x^2-3^2)=(x-3)(x+3)#........(4)

Substitute (4) into (1) giving:

#9/((x-3)(x+3)) - 5/(x+3)#

Now both denominators have something in common so we can combine them

#(9-5(x-3))/((x+3)(x-3))#

#(9-5x+15)/((x+3)(x-3))#

#((-5x)+24)/ ((x+3)(x-3))# .................(5)

but #(-5x+24) = -(5x-24)#...............(6)

substituting (6) into (5) and rewriting the expression gives:

#- (5x-24)/(x^2-9)#

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