Equations with 3 unknowns

Question

I have this equation/problem: Find the value of A and B if 2/(x−5)(x+3) = A /(x−5) + B/(x+3)

How can I solve/approach this? Thank you for your advice.

Regards Lisa

Answer 1

This question is about decomposing into partial fractions.

To start with we know the l.h.s is equal to the r.h.s, and therefore we can write the following $$ \frac{A}{x-5} +\frac{B}{x+3} = \frac{A(x+3) + B(x-5)}{(x-5)(x+3)} =\frac{(A+B)x + (3A-5B)}{(x-5)(x+3)} $$ which we can set to be $$ \frac{(A+B)x + (3A-5B)}{(x-5)(x+3)} = \frac{2}{(x-5)(x+3)} $$ now we can compare coefficients.

First thing to notice, we don't have a dependency on $x$, this implies $$ A + B = 0 \implies A = -B $$ now we have $$ 2 = 3A - 5B = 3(-B) - 5B = -8B \implies B = -\frac{2}{8} = -\frac{1}{4} $$ and finally $$ A = \frac{1}{4} $$

Answer 2

$\dfrac2{(x-5)(x+3)}=\dfrac A {x-5} + \dfrac B {x+3} \iff 2 = A(x+3)+B(x-5)=(A+B)x+(3A-5B).$

So set $0=A+B$ and $2=3A-5B. $ Can you solve for $A$ and $B$ now?