If frac{1}{(3-5 x)(2+3 x)}=frac{A}{3-5 x}+fra | vedclass

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(B) Given,$\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$ Equating the numerators: $1 = A(2+3x) + B(3-5x)$ $1 = (2A + 3B) + x(3A - 5B)$ Com

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$\frac{8}{19}$

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(B) Given,$\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$ Equating the numerators: $1 = A(2+3x) + B(3-5x)$ $1 = (2A + 3B) + x(3A - 5B)$ Comparing the coefficients of $x$ and the constant terms on both sides: $3A - 5B = 0$  ...$(i)$ $2A + 3B = 1$  ...$(ii)$ Multiplying equation $(i)$ by $3$ and equation $(ii)$ by $5$: $9A - 15B = 0$ $10A + 15B = 5$ Adding these two equations: $19A = 5 \implies A = \frac{5}{19}$ Substituting $A$ in equation $(i)$: $3(\frac{5}{19}) - 5B = 0 \implies 5B = \frac{15}{19} \implies B = \frac{3}{19}$ Therefore,$A+B = \frac{5}{19} + \frac{3}{19} = \frac{8}{19}$

If $\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$,then $A+B$ is equal to

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