If #a/(x-2)+b/(x+3)=(2x+11)/(x^2+x-6)#, then (a,b) = ?

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Answer 1 · 2017-12-07T09:52:24

#3/(x-2)-1/(x+3)#

This can be solvec by using partial fractions.

We know that #(2x+11)/(x^2+x-6)=a/(x-2)+b/(x+3)#

We can add these two fractions to get:
#a/(x-2)+b/(x+3)=(a(x+3))/((x-2)(x+3))+(b(x-2))/((x-2)(x+3))#
#=(a(x+3)+b(x-2))/((x+3)(x-2))#

By multiplying by #(x-3)(x+2)#, we get:

#2x+11=a(x+3)+b(x-2)#

First, we will use #x=2# as this will cancel out #b# and allow us to work out #a#.

#2(2)+11=a(2+3)+b(2-2)#

#4+11=a(5)+b(0)#

#15=5a#

#a=15/5=3#

Now, we do the same for #x=-3#

#2(-3)+11=a(-3+3)+b(-3-2)#

#-6+11=a(0)+b(-5)#

#5=-5b#

#b=-5/5=-1#

#a=3,b=-1#

#(2x+11)/(x^2+x-6)=3/(x-2)+(-1)/(x+3)#

#=3/(x-2)-1/(x+3)#