Mathematics

If $$\displaystyle \frac{3x+4}{x^2-3x+2}=\frac{A}{x-2}-\frac{B}{x-1}$$, then $$(A,B)$$=


ANSWER

$$(10, 7)$$


SOLUTION
$$\displaystyle \frac{3x+4}{x^2-3x+2}=\frac{A}{x-2}-\frac{B}{x-1}$$            ....(1)

Consider,$$\displaystyle \frac { 3x+4 }{ x^{ 2 }-3x+2 } $$
Resolving into partial fractions
$$\displaystyle \frac { 3x+4 }{ (x-1)(x-2) } =\frac { A }{ x-2 } +\frac { B }{ x-1 } $$     ....(2)
$$3x+4=A(x-1)+B(x-2)$$
$$ 3x+4=(A+B)x+(-A-2B)$$
$$\Rightarrow A+B=3,-A-2B=4$$
Solving these equations, we get
$$B=-7, A=10$$
Put these values in (2)
$$\displaystyle \frac { 3x+4 }{ (x-1)(x-2) } =\frac { 10 }{ x-2 } -\frac { 7 }{ x-1 } $$
Comparing this with (1), we get
$$(A,B) = (10,7)$$
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