How can you evaluate #(x)/(x^2-36) - (6)/(x^2-36)#?

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Answer 1 · 2015-07-22T15:58:47

#=color(blue)(1/(x+6)#

#x/(x^2-36) - 6/(x^2-36)#

The denominators of both terms being common we can simply add the numerators:

#=(x-6)/(x^2-36) #

Now, as per property:
#color(blue)(a^2-b^2= (a+b)(a-b)#

Expressing the denominator in the same manner:
#color(blue)((x^2-36) = (x+6)(x-6)#

The expression becomes:

#=cancel(x-6)/(color(blue)( (x+6)cancel(x-6))#

#=color(blue)(1/(x+6)#