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Question:

$\frac{a}{(a x-1)}+\frac{b}{(b x-1)}=(a+b), \quad\left(x \neq \frac{1}{a}, \frac{1}{b}\right)$

 

Solution:

$\frac{a}{(a x-1)}+\frac{b}{(b x-1)}=(a+b)$

$\Rightarrow\left[\frac{a}{(a x-1)}-b\right]+\left[\frac{b}{(b x-1)}-a\right]=0$

$\Rightarrow \frac{a-b(a x-1)}{a x-1}+\frac{b-a(b x-1)}{b x-1}=0$

$\Rightarrow \frac{a-a b x+b}{a x-1}+\frac{a-a b x+b}{b x-1}=0$

$\Rightarrow(a-a b x+b)\left[\frac{1}{(a x-1)}+\frac{1}{(b x-1)}\right]=0$

$\Rightarrow(a-a b x+b)\left[\frac{(b x-1)+(a x-1)}{(a x-1)(b x-1)}\right]=0$

$\Rightarrow(a-a b x+b)\left[\frac{(a+b) x-2}{(a x-1)(b x-1)}\right]=0$

$\Rightarrow(a-a b x+b)[(a+b) x-2]=0$

$\Rightarrow a-a b x+b=0$ or $(a+b) x-2=0$

$\Rightarrow x=\frac{(a+b)}{a b}$ or $x=\frac{2}{(a+b)}$

Hence, the roots of the equation are $\frac{(a+b)}{a b}$ and $\frac{2}{(a+b)}$.

 

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