Solve each of the following quadratic equations:
Question:

Solve each of the following quadratic equations:

(i) $\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}, \quad x \neq 1,-5$

(ii) $\frac{1}{2 x-3}+\frac{1}{x-5}=1 \frac{1}{9}, \quad x \neq \frac{3}{2}, 5$

Solution:

(i)

$\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}, \quad x \neq 1,-5$

$\Rightarrow \frac{x+5-x+1}{(x-1)(x+5)}=\frac{6}{7}$

$\Rightarrow \frac{6}{x^{2}+4 x-5}=\frac{6}{7}$

$\Rightarrow x^{2}+4 x-5=7$

$\Rightarrow x^{2}+4 x-12=0$

$\Rightarrow x^{2}+6 x-2 x-12=0$

$\Rightarrow x(x+6)-2(x+6)=0$

$\Rightarrow(x+6)(x-2)=0$

$\Rightarrow x+6=0$ or $x-2=0$

$\Rightarrow x=-6$ or $x=2$

Hence, −6 and 2 are the roots of the given equation.

(ii)

$\frac{1}{2 x-3}+\frac{1}{x-5}=1 \frac{1}{9}$

$\Rightarrow \frac{1}{2 x-3}+\frac{1}{x-5}=\frac{10}{9}$

$\Rightarrow \frac{(x-5)+(2 x-3)}{(2 x-3)(x-5)}=\frac{10}{9}$

$\Rightarrow \frac{3 x-8}{2 x^{2}-3 x-10 x+15}=\frac{10}{9}$

$\Rightarrow \frac{3 x-8}{2 x^{2}-13 x+15}=\frac{10}{9}$

$\Rightarrow 27 x-72=20 x^{2}-130 x+150$

$\Rightarrow 20 x^{2}-157 x+222=0$

$\Rightarrow 20 x^{2}-120 x-37 x+222=0$

$\Rightarrow 20 x(x-6)-37(x-6)=0$

$\Rightarrow(20 x-37)(x-6)=0$

$\Rightarrow 20 x-37=0$ or $x-6=0$

$\Rightarrow x=\frac{37}{20}$ or $x=6$