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I've found the common denominators and then played around with the stuff in the numerator, but the question asks specifically $A + B$. Is that even possible to find since we have A - B on the left hand side?

So I got $A(x+2) - B(x-3) = 3x + 16$

Just not sure if you can solve for an $A + B$. Thanks for any help.

Jessica Tiberio
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JCase
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    Hint: You can reorder the equation you obtained to $$ [A -B]x + [2A + 3B] = 3x + 16 $$ – Matti P. May 23 '18 at 12:26
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    From A(x+ 2)- B(x- 3)= 3x+ 16, we have Ax+ 2A- Bx+ 3B= (A- B)x+ (2A+ 3B)= 3x+ 16. In order for that to be true for all x, we must have A- B= 3 and 2A+ 3B= 16. Solve those equations for A and B. Once you have found A and B separately, add to get A+ B. – user247327 May 23 '18 at 12:28
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    Ok (assuming that the denom is correctly written) : $Ax + 2A - Bx + 3B = (A-B)x +(2A+3B)=3x + 16$. This means : $A-B=3$ and $2A+3B=16$. – Mauro ALLEGRANZA May 23 '18 at 12:31
  • It should be $x^2-x-6$ if the equation is supposed to hold for all $x$. In that case, you can just put $x=5/2$, and multiply the whole equation by $4-(5/2)^2$ and you get directly $A+B$. What is $5/2$? The solution of $x^2-x-6=-(x^2-4)$ that doesn't make the denominators vanish. –  May 23 '18 at 12:34

1 Answers1

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Hint: You write,

I got $$ A(x+2) - B(x-3) = 3x + 16. $$

The left hand side can be simplified into the form $$ \text{something}~x + \text{somethingElseWithoutX} $$

I'll get you started: $A(x + 2) = Ax + 2A$.

Try doing that, and see what you learn about $A$ and $B$.

John Hughes
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  • Thank you for your help!!!

    So you were saying that you can rearrange the equation such as (A-B)x + (2A + 3B) = 3x + 16....

    Then, A-B must equal 3, and 2A + 3B must equal 16.

    So, A-B=3, or A=3+B. Then, Substitute into the other equation. 2(3+B) + 3B = 16. We have B=2. If B=2, A=5.

    Amazing.

     – JCase May 23 '18 at 13:10