Question
Solve: $\frac{1}{2 x-3}+\frac{1}{x-5}=1 \frac{1}{9}, x \neq \frac{3}{2}, 5$
✓
Answer
The given equation is:
$\frac{1}{2 x-3}+\frac{1}{x-5}=\frac{10}{9}$
$\Rightarrow \frac{x-5+2 x-3}{(2 x-3)(x-5)}=\frac{10}{9}$
$\Rightarrow \frac{3 x-8}{(2 x-3)(x-5)}=\frac{10}{9}$
$\Rightarrow 27 x-72=10[(2 x-3)(x-5)] \text { ( By cross multiplication method) }$
$\Rightarrow 27 x-72=10\left[2 x^2-10 x-3 x+15\right]$
$\Rightarrow 27 x-72=10\left[2 x^2-13 x+15\right]$
$\Rightarrow 27 x-72=20 x^2-130 x+150$
$\Rightarrow 20 x^2-157 x+222=0$
Here $, a =20, b=-157, c =222$
Therefore, by quadratic formula we have:
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{157 \pm \sqrt{(-157)^2-4(20)(222)}}{40}$
$\Rightarrow x=\frac{157 \pm \sqrt{24649-17760}}{40}$
$\Rightarrow x=\frac{157 \pm \sqrt{6889}}{40}$
$\Rightarrow x=\frac{157 \pm 83}{40}$
$\Rightarrow x=\frac{157+83}{40} \text { or } x=\frac{157-83}{40}$
$\Rightarrow x=6 \text { or } x=\frac{37}{20}$
$\frac{1}{2 x-3}+\frac{1}{x-5}=\frac{10}{9}$
$\Rightarrow \frac{x-5+2 x-3}{(2 x-3)(x-5)}=\frac{10}{9}$
$\Rightarrow \frac{3 x-8}{(2 x-3)(x-5)}=\frac{10}{9}$
$\Rightarrow 27 x-72=10[(2 x-3)(x-5)] \text { ( By cross multiplication method) }$
$\Rightarrow 27 x-72=10\left[2 x^2-10 x-3 x+15\right]$
$\Rightarrow 27 x-72=10\left[2 x^2-13 x+15\right]$
$\Rightarrow 27 x-72=20 x^2-130 x+150$
$\Rightarrow 20 x^2-157 x+222=0$
Here $, a =20, b=-157, c =222$
Therefore, by quadratic formula we have:
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{157 \pm \sqrt{(-157)^2-4(20)(222)}}{40}$
$\Rightarrow x=\frac{157 \pm \sqrt{24649-17760}}{40}$
$\Rightarrow x=\frac{157 \pm \sqrt{6889}}{40}$
$\Rightarrow x=\frac{157 \pm 83}{40}$
$\Rightarrow x=\frac{157+83}{40} \text { or } x=\frac{157-83}{40}$
$\Rightarrow x=6 \text { or } x=\frac{37}{20}$
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