Question
Solve: $\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=2, x \neq-\frac{1}{2}, 1$
✓
Answer
Given
$\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=0$
Let $\frac{x-1}{2 x+1}$ be $y$ so $\frac{2 x+1}{x-1}=\frac{1}{y}$
$\therefore$ Substituting this value
$y+\frac{1}{y}=2 \text { or } \frac{y^2+1}{y}=2$
or $y^2+1=2 y$
or $y^2-2 y+1=0$
$\text { or }(y-1)^2=0$
Putting $y=\frac{x-1}{2 x+1}$,
$\frac{x-1}{2 x+1}=1 \text { or } x-1=2 x+1$
$\text { or } x=-2$
$\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=0$
Let $\frac{x-1}{2 x+1}$ be $y$ so $\frac{2 x+1}{x-1}=\frac{1}{y}$
$\therefore$ Substituting this value
$y+\frac{1}{y}=2 \text { or } \frac{y^2+1}{y}=2$
or $y^2+1=2 y$
or $y^2-2 y+1=0$
$\text { or }(y-1)^2=0$
Putting $y=\frac{x-1}{2 x+1}$,
$\frac{x-1}{2 x+1}=1 \text { or } x-1=2 x+1$
$\text { or } x=-2$
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