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 does not equal -1 and x does not equal 2. Give your answer as the ordered pair.
+118673 Your answer is better than mine Alan. it is better explained.
I rarely touch questions that are presented so poorly. Maybe that is why Lightning reposted.
Or maybe this question was posted by someone else in his class.
Ref for others : https://web2.0calc.com/questions/help_51906
+1245 Multiplying both sides by $(x - 2)(x + 1) = x^2 - x - 2$, we get \begin{align*} x + 7 &= A(x + 1) + B(x - 2) \\ &= Ax + Bx + A - 2B \\ &= (A + B)x + (A - 2B). \end{align*} Comparing the coefficients on both sides, we obtain the system of equations \begin{align*} A + B &= 1, \\ A - 2B &= 7. \end{align*} Subtracting the second equation from the first equation, we get $3B = 1 - 7 = -6$, so $B = -2$. Then from the first equation, $A = 1 - B = 3$. Hence, \[\frac{x + 7}{x^2 - x - 2} = \frac{3}{x - 2} - \frac{2}{x + 1},\] and our answer is \((A,B) = \boxed{(3,-2)}\).
