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# Help! Algebra

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

Aug 17, 2022

#1
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Hi Guest!

This is a partial fraction question.

First, we factor the denominator:

$$\dfrac{x-7}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$$

Then, we multiply both sides by $$(x-2)(x+1)$$ to get rid of the denominators, and we know that $$x \neq -1,2$$

So: $$x-7=A(x+1)+B(x-2)$$

Next, we expand the right-hand side, and collect like terms:

$$x-7=Ax+A+Bx-2B \\ x-7 = (A+B)x+(A-2B)$$

Now, we equate the coefficients of x, and the constant:

$$1=A+B \\ -7=A-2B \\$$

This is a system of two linear equations.

We subtract the second from the first equation to get:

$$1-(-7)=A+B-(A-2B) \\ \iff 8 =3B \\ \therefore B=\dfrac{8}{3}$$

Now, substituting in the first equation:

$$1=A+\dfrac{8}{3} \implies A=-\dfrac{5}{3}$$

Therefore, $$(A,B) = (-\dfrac{5}{3},\dfrac{8}{3})$$

I hope this helps :)!

Aug 18, 2022
#2
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Thanks!!!!!

Guest Aug 18, 2022