Here's the question you clicked on:
geerky42
If \(\dfrac{x-8}{x+3}+\dfrac{x+3}{x-8}\) is equal to the mixed number \(A \dfrac{B}{(x+3)(x-8)}\), then B = ...
A. 64 B. 8 C. 121 D. 24 E. 9
So far, I have B + A(x+3)(x-8) = (x+3)²+(x-8)² I don't know what to do next...
Apparently, I need second equation somehow.
OK, so we add the fractions:\[\frac{ x-8 }{ x+3 }+\frac{ x+3 }{ x-8 }=\frac{ (x-8)^2+(x+3)^2 }{ (x+3)(x-8) }\]\[=\frac{ x^2-16x+64+x^2+6x+9 }{ (x+3)(x-8) }=\frac{ 2x^2-10x+73 }{ x^2-5x-24 }=\] From this it is clear that A=2. Now we have to get B!
The last fraction is:\[\frac{ 2x^2-10x+73 }{ x^2-5x-24 }=\frac{ 2x^2-10x-48+121 }{ x^2-5x-24 }=\frac{ 2(x^2-5x-24)+121 }{ x^2-5x-24 }\]\[=2\frac{ 121 }{ x^2-5x-24 }=2\frac{ 121 }{ (x+3)(x-8) }\]So B = 121.