Not entirely sure what the question is supposed to say, so here's my best guess.
First, find the partial fraction decomposition of
[tex]\dfrac{x-9}{x^2-3x-18}[/tex]
This is equal to
[tex]\dfrac{x-9}{(x-6)(x+3)}=\dfrac a{x-6}+\dfrac b{x+3}[/tex]
Multiply both sides by [tex](x-6)(x+3)[/tex], so that
[tex]x-9=a(x+3)+b(x-6)[/tex]
Notice that if [tex]x=6[/tex], the term involving [tex]b[/tex] vanishes, so that
[tex]6-9=a(6+3)\implies a=-\dfrac13[/tex]
Then if [tex]x=-3[/tex], the term with [tex]a[/tex] vanishes and we get
[tex]-3-9=b(-3-6)\implies b=\dfrac43[/tex]
So we have
[tex]\dfrac{x-9}{x^2-3x-18}=-\dfrac1{3(x-6)}+\dfrac4{3(x+3)}[/tex]
I think the final answer is supposed to be [tex]a+b[/tex], so you end up with 1.
