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# help

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What fraction of the form $$\frac{A}{x + 3}$$ can be added to $$\frac{6x}{x^2 + 2x - 3}$$ so that the result reduces to a fraction of the form $$\frac{B}{x - 1}$$? Here A and B are real numbers. Give the value of A as your answer.

Mar 10, 2019

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$$\dfrac{A}{x+3}+\dfrac{6x}{x^2+2x-3}=\dfrac{B}{x-1}\\ \dfrac{A}{x+3}+\dfrac{6x}{(x+3)(x-1)}=\dfrac{B}{x-1}\\ \dfrac{A(x-1)+6x}{(x+3)(x-1)}=\dfrac{B(x+3)}{(x+3)(x-1)}$$

$$(A+6)x-A=Bx + 3B$$

$$A+6 = B\\ -A = 3B\\ \text{Solve these two equations for }A, B$$

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Mar 10, 2019