Question

Solve for x :

${\displaystyle \frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9},X\ne \frac{3}{2},5}$

Answer 1

We have

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

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

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

⇒9(3x − 8) = 10(2x² − 13x + 15)

⇒27x − 72 = 20x² − 130x + 150

⇒20x² − 157x + 222 = 0

⇒20x² −  120x − 37x + 222 = 0

⇒20x(x − 6) − 37(x − 6) = 0

⇒(20x − 37)(x − 6) = 0

${\displaystyle \Rightarrow x=\frac{37}{20},6}$