Question

Solving $\frac{a}{bx}-\frac{b}{ax}=a^2-b^2$ gives $x$ equal to

• A. $1$
• B. $a+b$
• C. $\frac{1}{ab}$
• D. $a^2{b}^2$

Answer 1

Answer: (C)

$\frac{1}{ab}$

The given equality can be solved as follows:

$\frac{a}{bx}-\frac{b}{ax}=a^2-b^2\Rightarrow \frac{(a\times ax)-(b\times bx)}{\left(bx\right)\left(ax\right)}=a^2-b^2\left(TakingLCM\right)\Rightarrow \frac{a^{2}x-b^{2}x}{ab{x}^2}=a^2-b^2\Rightarrow \frac{x(a^2-b^2)}{ab{x}^2}=a^2-b^2$

$\Rightarrow \frac{x(a^2-b^2)}{ab{x}^2(a^2-b^2)}=1\Rightarrow \frac{1}{abx}=1\Rightarrow x=\frac{1}{ab}$

Hence, $x=\frac{1}{ab}$.