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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.

Aug 17, 2018

#1
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A partial fractions problem: .

Aug 18, 2018
#2
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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

Melody  Aug 18, 2018
#3
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I didn't double post this one. I would have realized the answer was already answered. Plus, I have a work from 9pm to 3am. So it coudn't be me.

Lightning  Aug 18, 2018
#4
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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 $$(A,B) = \boxed{(3,-2)}$$.

Aug 19, 2018