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# help Algebra

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There are values A and B such that

(Bx - 11)/(x^2 - 7x + 10) = A/(x - 2) + 8/(x - 5)
Find A+B.

Nov 12, 2021

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$$\frac{Bx-11}{x^2-7x+10}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\cdot(x-2)(x-5)\ =\ \Big(\frac{A}{x-2}+\frac{8}{x-5}\Big)\cdot(x-2)(x-5)\\~\\ Bx-11=\frac{A(x-2)(x-5)}{x-2}+\frac{8(x-2)(x-5)}{x-5}\\~\\ Bx-11\ =\ A(x-5)\ +\ 8(x-2)$$

Assuming the previous equation is true for all values of  x ,  it must be true when  x = 0 , so we can say:

$$B(0)-11\ =\ A(0-5)+8(0-2)\\~\\ -11\ =\ A(-5)-16\\~\\ 5\ =\ A(-5)\\~\\ A\ =\ -1$$

And it must also be true when  x = 5  so we can say:

$$B(5)-11\ =\ A(5-5) + 8(5 - 2)\\~\\ 5B-11\ =\ 24\\~\\ 5B\ =\ 35\\~\\ B\ =\ 7$$

So   A + B   =   -1 + 7   =   6

(BTW even if the 8 and the B are swapped you can still apply this same method to solve it!)

Nov 12, 2021