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Find constants A and B such that (3x - 17)/((x - 3)(x - 7) = A/(x - 3) + B/(x - 7).

May 31, 2020

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(3x - 17)/((x - 3)(x - 7) = A/(x - 3) + B/(x - 7)

Rewrite each fraction with the denominator:  (x - 3)(x - 7)

[ A/(x - 3) ] · [ (x - 7)/(x - 7) ]  =  [ A · (x - 7) ] / [ (x - 3)(x - 7) ]  =  [ Ax - 7A ] / [ (x - 3)(x - 7) ]

[ B(x - 7) ] · [ (x - 3)/(x - 3) ]  =  [ B · (x - 3) ] / [ (x - 3)(x - 7) ]  =  [ Bx - 3B ] / [ (x - 3)(x - 7) ]

[ Ax - 7A ] / [ (x - 3)(x - 7) ]  +  [ Bx - 3B ] / [ (x - 3)(x - 7) ]   =  [ Ax - 7A + Bx - 3B ] / [ (x - 3)(x - 7) ]

Comparing the original numerator  ( 3x - 17 )  to this numerator  ( Ax - 7A + Bx - 3B )

3x - 17  =  Ax - 7A + Bx - 3B

3x - 17  =  (Ax + Bx) + (- 7A - 3B)

Setting the x-terms equal to each other:       3x  =  Ax + Bx   --->     3  =  A + B

Setting the constants equal to each other:  -17  =  -7A - 3B   --->   17  =  7A + 3B

Solving for A:   A + B  =  3   --->   -A -    B  =  -3   --->   -3A - 3B  =  -9

7A + 3B  =  17   --->   7A + 3B  =  17

Solving:                                                                          4A           =  8     --->     A  =  2

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

May 31, 2020