Question

Solve for $x$ : $\frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9},x\ne \frac{3}{2},5$

Answer 1

$\frac{1}{2x-3}+\frac{1}{x-5}=\frac{10}{9}\Rightarrow \frac{x-5+2x-3}{(2x-3)(x-5)}=\frac{10}{9}\Rightarrow \frac{3x-8}{2x(x-5)-3(x-5)}=\frac{10}{9}\Rightarrow \frac{3x-8}{2x^2-10x-3x+15}=\frac{10}{9}\Rightarrow \frac{3x-8}{2x^2-13x+15}=\frac{10}{9}\Rightarrow 27x-72=20x^2-130x+150\Rightarrow 27x+130x=20x^2+72+150\Rightarrow 157x=20x^2+222\Rightarrow 20x^2+222-157=0\Rightarrow 20x^2-157x+222=0$

It is i form $a{x}^2+bx+c=0$

So by applying ShreedharAcharya formula

$\frac{-\left(157\right)\pm \sqrt{(187{)}^2-4\times 20\times 222}}{2\times 20}=\frac{157\pm \sqrt{24649-17760}}{40}=\frac{157\pm \sqrt{6889}}{40}=\frac{157+85}{40}or\frac{187-85}{40}=6or1.85$