question_answer

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

Answer:

Given, \[\frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9}\]

\[\frac{x-5+2x-3}{(2x-3)(x-5)}=\frac{10}{9}\]

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

\[9(3x-8)=10(2{{x}^{2}}-13x+15)\]

\[27x-72=20{{x}^{2}}-130x+150\]

\[20{{x}^{2}}-157x+222=0\]

\[20{{x}^{2}}-120x-37x+222=0\]

\[20x(x-6)-37(x-6)=0\]

\[(x-6)(20x-37)=0\]

Either \[x-6=0\] or \[20x-37=0\]

\[\Rightarrow \]                          \[x=6\]\[x=\frac{37}{20}\]