If frac{2x + 3}{(x + 1)(x - 3)} = frac{a}{x + | vedclass

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

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$2$

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(B) Given the partial fraction decomposition: $\frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3}$.Multiplying both sides by $(x + 1)(x - 3)$,we get: $2x + 3 = a(x - 3) + b(x + 1)$.Setting $x = -1$: $2(-1) + 3 = a(-1 - 3)$ $\Rightarrow 1 = -4a$ $\Rightarrow a = -\frac{1}{4}$.Setting $x = 3$: $2(3) + 3 = b(3 + 1)$ $\Rightarrow 9 = 4b$ $\Rightarrow b = \frac{9}{4}$.Therefore,$a + b = -\frac{1}{4} + \frac{9}{4} = \frac{8}{4} = 2$.

If $\frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3}$,then the value of $a + b$ is:

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