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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.

Guest Aug 17, 2018
 #1
avatar+27133 
+2

A partial fractions problem:

 

.

Alan  Aug 18, 2018
 #2
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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

Melody  Aug 18, 2018
 #3
avatar+803 
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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
avatar+803 
+1

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 \((A,B) = \boxed{(3,-2)}\).

Lightning  Aug 19, 2018

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