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

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(D) Given the equation: $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$Since $x^2 - 3x + 2 = (x - 2)(x - 1)$,we can write:$\frac{3x

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$(a)$ and $(c)$ both

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(D) Given the equation: $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$Since $x^2 - 3x + 2 = (x - 2)(x - 1)$,we can write:$\frac{3x + a}{(x - 2)(x - 1)} = \frac{A(x - 1) - 10(x - 2)}{(x - 2)(x - 1)}$Equating the numerators:$3x + a = A(x - 1) - 10(x - 2)$$3x + a = Ax - A - 10x + 20$$3x + a = (A - 10)x + (20 - A)$Comparing the coefficients of $x$ and the constant terms:$3 = A - 10 \implies A = 13$$a = 20 - A \implies a = 20 - 13 = 7$Thus,both $a = 7$ and $A = 13$ are correct. Therefore,the correct option is $(d)$.

If $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$,then

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