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# Halp

0
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5

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 $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

Jan 15, 2020

### 5+0 Answers

#1
0

(A,B) = (5,-4).

Jan 15, 2020
#2
0

That did not work I think it is wrong

Jan 15, 2020
#3
0

Help me

Jan 15, 2020
#4
+24402
+2

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 $$(A,B)$$.

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

$$\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{x + 7}{x^2 - x - 2} } &=& \mathbf{ \dfrac{A}{x - 2} + \dfrac{B}{x + 1} } \quad | \quad x^2 - x - 2=(x-2)(x+1) \\\\ &\dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x-2)(x+1) \\\\ &\dfrac{(x + 7)(x-2)(x+1)}{(x-2)(x+1)} &=& \dfrac{A(x-2)(x+1)}{(x-2)} + \dfrac{B(x-2)(x+1)}{(x+1)} \\\\ & x + 7 &=& A(x+1) + B(x-2) \\\\ \mathbf{x=-1:}& -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& A\times 0 -3B \\ & 3B &=& -6 \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& -2 \\\\ \mathbf{x=2:}& 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3A +B\times 0\\ & 3A &=& 9 \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& 3 \\ \hline \end{array}$$

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

check:

$$\begin{array}{|rcll|} \hline \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad A=3,\ B=-2 \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3}{x - 2} - \dfrac{2}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3x+3-2x+4}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{x+7}{(x-2)(x+1)} \ \checkmark \\ \hline \end{array}$$

Jan 15, 2020
#5
+1

thanks

Jan 15, 2020