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Find the constants A and B that makes the given equation true. How do I solve for this?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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\[\frac{3x+15}{x^2-4x-5}=\frac{a}{x-50}+\frac{b}{x+1}\]
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@zepdrix @dan815
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I don't know where to start with this equation..

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@jim_thompson5910
Victoriasushchik
  • Victoriasushchik
Multiply to get the denominators equal first. You are on the right track. 2x-9/(x^2-x-6) = A/(x-3) + B/(x+2) Multiply A by (x+2)/(x+2) and B by (x-3)/(x-3). Now all have the same denominator, so you can ignore the denominator when solving for A and B. Now you have... 2x-9 = A(x+2) + B(x-3) Distribute... 2x-9 = Ax + 2A + Bx - 3B Separate into two equations, one for X and one for your constants. 2x = Ax + Bx -9 = 2A - 3B Solve this system of equations. You can drop the x's out of the first equation because there is one in every term. 2 = A + B -9 = 2A - 3B Multiply the first equation by 3 and add to the second to cancel out the B's. 6 = 3A + 3B Add -9 = 2A - 3B + ___________ -3 = 5A A = -3/5. Plug this back into any of the two equations. 2 = A + B ===> 2 = -3/5 + B 13/5 = B So, A = -3/5. and B = 13/5.
Victoriasushchik
  • Victoriasushchik
Is This Helpful??
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wOw thank you so much! I was taking my time to process the stuff. U r a lifesaver :D
Victoriasushchik
  • Victoriasushchik
your welcome!!!
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I really appreciate your time and effort to help me!
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yo @jim_thompson5910 thanks for coming by
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Also I mistyped the problem but I understand how to solve
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it suppose to be (x-5) not (x-50)
jim_thompson5910
  • jim_thompson5910
Here is another route 2x-9 = A(x+2) + B(x-3) 2(-2)-9 = A(-2+2) + B(-2-3) .... plug in x = -2 -4-9 = A*0+B(-5) -13 = -5B -13/(-5) = B B = 13/5 Now use this to find A. You'll find that A = -3/5
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Oh okey so just use x to get ride of one of the variables. Gotcha!
jim_thompson5910
  • jim_thompson5910
@Victoriasushchik has the more effective method though

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