Question

$\frac{2x+1}{(x+a)(bx-1)}=\frac{5}{x+a}-\frac{3}{bx-1}\Rightarrow (a,b)=$

• A. $(2,-1)$
• B. $(-2,-1)$
• C. $(-2,1)$
• D. (2, 1)

Answer 1

The correct option is C $(-2,1)$

LHS $\frac{2x+1}{(x+a)(bx-1)}$ ;RHS $=\frac{5(bx-1)-3(x+a)}{(x+a)(bx-1)}$
LHS = RHS $2x+1=5(bx-1)-3(x+a)$ ------------(1)
Since this is the equation it must be true for any value of x
$\Rightarrow$ let $x=0$ we get from (1)
we get $1=5(-1)-3\left(a\right)$
$6=-3\left(a\right)$ $\Rightarrow$ $a=-2$ -----------(1)
$5=5(2b-1)$
$\Rightarrow 2b=2$
$b=1$
So, $(a,b)=(-2,1)$