Question

please solve and answer please​

Answer 1

$\implies \dfrac{ax + b}{ {(3x + 4)}^{2} } = \dfrac{1}{ {(3x + 4)} } - \dfrac{3}{ {(3x + 4)}^{2} }$

Now RHS is :-

$\implies \dfrac{1}{ {(3x + 4)} } - \dfrac{3}{ {(3x + 4)}^{2} }$

LCM of (3x+4) and (3x+4)² is = (3x+4)²

$\implies \dfrac{3x + 4 - 3}{{ {(3x + 4)}^{2} }}$

$\implies \dfrac{3x + 1}{{ {(3x + 4)}^{2} }}$

therefore LHS :-

$\implies \dfrac{3x + 1}{{ {(3x + 4)}^{2} }}$

Now RHS is :-

$\implies \dfrac{ax + b}{ {(3x + 4)}^{2} }$

Now , we know LHS = RHS .

$\implies \dfrac{ax + b}{ {(3x + 4)}^{2} } = \dfrac{3x + 1}{{ {(3x + 4)}^{2} }}$

since denominator of both LHS and RHS are equal , so we will compare numerator.

$\implies {ax + b} = {3x + 1}$

Therefore we get :-

$\implies \: b = 1$

$\implies \: ax = 3x$

cancel out x from both side.

$\implies \: a= 3$

Answer :

b = 1 and a = 3