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

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(D) For the first partial fraction: $\frac{1}{(3x+1)(x-2)} = \frac{A}{3x+1} + \frac{B}{x-2}$.By partial fraction decomposition,$1 = A(x-2) + B(3x+1)$.

Vedclass · Accepted Answer

$A+3B=0, A:C=3:2, B:D=1:3$

Vedclass · Answer

(D) For the first partial fraction: $\frac{1}{(3x+1)(x-2)} = \frac{A}{3x+1} + \frac{B}{x-2}$.By partial fraction decomposition,$1 = A(x-2) + B(3x+1)$.Setting $x=2$,$1 = B(6+1) \Rightarrow B = \frac{1}{7}$.Setting $x=-\frac{1}{3}$,$1 = A(-\frac{1}{3}-2)$ $\Rightarrow 1 = A(-\frac{7}{3})$ $\Rightarrow A = -\frac{3}{7}$.Thus,$A+3B = -\frac{3}{7} + 3(\frac{1}{7}) = 0$.For the second partial fraction: $\frac{x+1}{(3x+1)(x-2)} = \frac{C}{3x+1} + \frac{D}{x-2}$.$x+1 = C(x-2) + D(3x+1)$.Setting $x=2$,$3 = D(7) \Rightarrow D = \frac{3}{7}$.Setting $x=-\frac{1}{3}$,$\frac{2}{3} = C(-\frac{7}{3}) \Rightarrow C = -\frac{2}{7}$.Now,$A:C = (-\frac{3}{7}) : (-\frac{2}{7}) = 3:2$ and $B:D = (\frac{1}{7}) : (\frac{3}{7}) = 1:3$.Therefore,$A+3B=0, A:C=3:2, B:D=1:3$.

If $\frac{1}{(3x+1)(x-2)}=\frac{A}{3x+1}+\frac{B}{x-2}$ and $\frac{x+1}{(3x+1)(x-2)}=\frac{C}{3x+1}+\frac{D}{x-2}$,then

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