The partial fraction decomposition of LaTeX: \frac{x-9}{x^2-3x-18} x − 9 x 2 − 3 x − 18 is LaTeX: \frac{A}{x-6}+\frac{B}{x+3} A x − 6 + B x + 3 . Find the numbers LaTeX: A\: A and LaTeX: B B . Then, find the sum LaTeX: A+B A + B , which is a whole number. Enter that whole number as your answer.

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.