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).

Guest Aug 17, 2022

#1**0 **

Hi Guest!

This is a partial fraction question.

First, we factor the denominator:

\(\dfrac{x-7}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}\)

Then, we multiply both sides by \((x-2)(x+1)\) to get rid of the denominators, and we know that \(x \neq -1,2\)

So: \(x-7=A(x+1)+B(x-2)\)

Next, we expand the right-hand side, and collect like terms:

\(x-7=Ax+A+Bx-2B \\ x-7 = (A+B)x+(A-2B)\)

Now, we equate the coefficients of x, and the constant:

\( 1=A+B \\ -7=A-2B \\\)

This is a system of two linear equations.

We subtract the second from the first equation to get:

\(1-(-7)=A+B-(A-2B) \\ \iff 8 =3B \\ \therefore B=\dfrac{8}{3}\)

Now, substituting in the first equation:

\(1=A+\dfrac{8}{3} \implies A=-\dfrac{5}{3}\)

Therefore, \((A,B) = (-\dfrac{5}{3},\dfrac{8}{3})\)

I hope this helps :)!

Guest Aug 18, 2022