If frac{13x+43}{2x^2+17x+30} = frac{A}{2x+5} | vedclass

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(A) Given the partial fraction decomposition: $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$ Combining the terms on the right side: $\f

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$8$

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(A) Given the partial fraction decomposition: $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$ Combining the terms on the right side: $\frac{13x+43}{2x^2+17x+30} = \frac{A(x+6) + B(2x+5)}{(2x+5)(x+6)}$ Since the denominators are equal,we equate the numerators: $13x + 43 = A(x+6) + B(2x+5)$ Expanding the right side: $13x + 43 = (A + 2B)x + (6A + 5B)$ Comparing the coefficients of $x$ and the constant terms,we get the system of equations: $A + 2B = 13$ ... $(i)$ $6A + 5B = 43$ ... $(ii)$ From $(i)$,$A = 13 - 2B$. Substituting this into $(ii)$: $6(13 - 2B) + 5B = 43$ $78 - 12B + 5B = 43$ $-7B = 43 - 78$ $-7B = -35 \Rightarrow B = 5$ Substituting $B = 5$ back into $(i)$: $A + 2(5) = 13 \Rightarrow A + 10 = 13 \Rightarrow A = 3$ Therefore,$A + B = 3 + 5 = 8$.

If $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$,then $A+B = $

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