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Find constants A and B such that (x + 7)/(x^2 + 3x + 2) = A/(x + 2) + B/(x + 1) for all x such that x ≠ -1 and x ≠ -2. Give your answer as the ordered pair (A,B).

 Dec 16, 2021
 #1
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(x + 7) / (x2 - x - 2)  =  A / (x - 2) + B / (x + 1)

 

Since  x2 - x - 2  =  (x - 2)(x + 1)

--->   (x + 7) / (x2 - x - 2)  =  (x + 7) / [ (x - 2)(x + 1) ]  =

 

(x + 7) / [ (x - 2)(x + 1) ]  =  A / (x - 2) + B / (x + 1)

 

Writing with the common denominator of  (x - 2)(x + 1)

(x + 7) / [ (x - 2)(x + 1) ]  =  [ A(x + 1) ] / [ (x - 2)(x + 1) ] + [B(x - 2) ] / [ (x - 2)(x + 1) ]

 

Since the denominators are equal, they can be ignored:

x + 7  =  A(x + 1) + B(x - 2)

x + 7  =  Ax + A + Bx - 2B

x + 7  =  Ax + Bx + A - 2B

Set the x-term on the left side equal to the x-terms on the right side:  x  =  Ax + Bx

   --->   x  =  (A + B)x   --->   1x  =  (A + B)x   --->   1  =  A + B

Set the number term on the left side equal to the number terms on the right side:  7  =  A - 2B

 

Solving:  A +   B  =  1   --->   x -1   --->   -A -  B  =  -1

                A - 2B  =  7                      --->    A - 2B  =  7

Adding down the columns:                         -3B  =  6

                                                                           B  =  -2

 

Since  B = -2:   A + B  =  1   --->   A + -2  =  1   --->   A  =  3

 

(-2, 3)

 

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 Dec 16, 2021
edited by AlgebraGuru  Dec 16, 2021
edited by AlgebraGuru  Dec 16, 2021

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