If frac{3x + 4}{x^2 - 3x + 2} = frac{A}{x - 2 | vedclass

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Question

(B) Given the partial fraction decomposition: $\frac{3x + 4}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{B}{x - 1}$.Multiplying both sides by $(x - 2)(x

Vedclass · Accepted Answer

$(10, 7)$

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{3x + 4}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{B}{x - 1}$.Multiplying both sides by $(x - 2)(x - 1)$,we get: $3x + 4 = A(x - 1) - B(x - 2)$.Expanding the right side: $3x + 4 = Ax - A - Bx + 2B = (A - B)x + (-A + 2B)$.Comparing the coefficients of $x$ and the constant terms:$A - B = 3$ (Equation $1$)$-A + 2B = 4$ (Equation $2$)Adding Equation $1$ and Equation $2$: $(A - B) + (-A + 2B) = 3 + 4 \Rightarrow B = 7$.Substituting $B = 7$ into Equation $1$: $A - 7 = 3 \Rightarrow A = 10$.Therefore,$(A, B) = (10, 7)$.

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

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