SOLUTION: Hi, I'm not sure which section to be asking these questions under, but i would really appreciate the help! :) Find A and B such that: 2x/(2x-3)(x+5)=A/2x-3+B/x+5

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Question 447019: Hi, I'm not sure which section to be asking these questions under, but i would really appreciate the help! :)
Find A and B such that: 2x/(2x-3)(x+5)=A/2x-3+B/x+5
Found 3 solutions by stanbon, Alan3354, sudhanshu_kmr:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
Find A and B such that: 2x/(2x-3)(x+5)=A/(2x-3)+B/x+5
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lcd = (2x-3)(x+5)
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(2x+0)/lcd = [A(x+5)-B(2x-3)]/lcd
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Equate the numerators:
2x+0 = (A-2B)x+(5A+3B)
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Solve:
A-2B = 2
5A+3B=0
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Multiply 1st by 5:
5A-10B = 10
5A+3B = 0
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Subtract and solve for "B":
13B = -10
B = -10/13
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Solve for "A":
A - 2B = 2
A -2(-10/13) = 2
A + 20/13 = 26/13
A = 6/13
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Cheers,
Stan H.
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Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Find A and B such that: 2x/(2x-3)(x+5)=A/2x-3+B/x+5
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2x/(2x-3)(x+5) = A/(2x-3)+ B/(x+5)
A(x+5) + B(2x-3) = 2x
Ax + 2Bx + 5A - 3B = 2x
A + 2B = 2
5A - 3B = 0
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A = 6/13
B = 10/13

Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!

2x/(2x-3)(x+5)=A/2x-3 + B/x+5

A/2x-3 + B/x+5 = [ A(x+5) + B(2X-3) ]/(2X-3)(X+5)
= [ X(A+2B) + 5A-3B ]/ (2X-3)(X+5)

now, compare both numerator and compare value of X and constant values..
A+2B = 2 , 5A-3B = 0
solving these equations, A = 6/ 13 , B = 10/13