Question

Solve:
(i). $\frac{1}{18-x}-\frac{1}{18+x}=\frac{1}{24}$ and $x>0$

Answer 1

$\frac{1}{18-x}-\frac{1}{18+x}=\frac{1}{24}$ and $x>0$

$\Rightarrow \frac{(18+x)-(18-x)}{(18-x)(18+x)}=\frac{1}{24}$

$\Rightarrow \frac{2x}{{18}^2-x^2}=\frac{1}{24}$

$\{\because (a-b\left)\right(a+b)=a^2-b^2\}$

$\Rightarrow 48x=324-x^2$

$\Rightarrow x^2+48x-324=0$

$\Rightarrow x^2+54x-6x-324=0$

$\Rightarrow x(x+54)-6(x+54)=0$

$\Rightarrow (x+54)(x-6)=0$

$\Rightarrow x+54=0$ or $x-6=0$

$\Rightarrow x=-54$ or $x=6$

Since $x>0$, so $x$ can’t be negative.

Hence, $x=6$.