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# Help!!

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Find constants $$A$$ and $$B$$ such that
$$\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$$
for all $$x$$ such that $$x \neq -1$$ and $$x\neq2$$. Give your answer as the ordered pair $$(A,B)$$.

Apr 16, 2020

#1
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Find constants $$A$$ and $$B$$ such that $$\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}$$
for all $$x$$ such that $$x\neq -1$$ and $$x\neq 2$$.
Give your answer as the ordered pair $$(A,\ B)$$.

$$\text{Let x^2 - x - 2 = (x-2)(x+1)}$$

$$\begin{array}{|rcll|} \hline \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad & | \quad \cdot (x-2)(x+1) \\\\ x + 7 &=& \dfrac{A(x-2)(x+1)}{x - 2} + \dfrac{B(x-2)(x+1)}{x + 1} \quad & | \quad x\ne2,\ x\ne -1 \\\\ x + 7 &=& A (x+1) + B(x-2) \\ \\ \hline x=-1: \quad -1 + 7 &=& A (-1+1) + B(-1-2) \\ \quad 6 &=& 0 -3B \\ \quad 6 &=& -3B \\ \quad -3B &=& 6 \quad | \quad :(-3) \\ \quad B &=& \frac{6}{-3} \\ \quad \mathbf{B} & \mathbf{=} & \mathbf{-2} \\ \hline x=2: \quad 2 + 7 &=& A (2+1) + B(2-2) \\ \quad 9 &=& 3A +0 \\ \quad 9 &=& 3A \\ \quad 3A &=& 9 \quad | \quad :3 \\ \quad A &=& \frac{9}{3} \\ \quad \mathbf{A} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}$$

$$\mathbf{(A,\ B) = (3,\ -2)}$$

Apr 16, 2020
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Can you go into more detail on why you specifically used the numbers it labeled as undefined?

Joshy1028  Apr 16, 2020
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Explanation:

It doesn't depend on the numbers labelled undefined

but it was the numbers that make one variable to be zero

For example in this question:

$$A(x+1)+B(x-2)$$ If x=-1 then 0A=0 so we cancelled one variable

and then x=2 to cancel B and find A

Guest Apr 16, 2020
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https://www.youtube.com/watch?v=BvGKVn-85jM Partial fraction's video, hope it helps!

Guest Apr 16, 2020