Find constants A and B such that
\[\frac{x-7}{x^2 -x-2}=\frac{A}{x-2} + \frac{B}{x+1}\]
for all x such that $x \neq -1$ and $x \neq 2$. Give your answer as the ordered pair (A,B).
+307 Answer: \((-\frac5 3,\frac8 3) \)
Solution:
The right side of the equation can be turned into \(A(x+1) + B(x-2)\over x^2 -x-2\). After that, you can multiply both sides of the equation to get \(x-7 = A(x+1) + B(x-2)\).
The coefficient of the x on the left, which is one, is equal to A + B, because the x term on the left side is made up of Ax + Bx. The constant term on the left side, -7, is equal to A-2B. This is because when expanding out the terms on the right, you would get Ax +\(A\)+Bx \(-2B\). From this information, you get two equations:
\(A+B=1\)
and
\(A-2B=-7\)
Subtracting the second equation from the first gives 3B = 8, which means that B = 8/3. Substituting this value into the first equation gives A + 8/3 = 1. Solving this gives A = 1-8/3 = -5/3.
Putting this into an ordered pair gives \((-\frac5 3,\frac8 3) \).
