Question
Solve:
$\left(\frac{3 x+1}{x+1}\right)+\left(\frac{x+1}{3 x+1}\right)=\frac{5}{2}$
$\left(\frac{3 x+1}{x+1}\right)+\left(\frac{x+1}{3 x+1}\right)=\frac{5}{2}$
✓
Answer
$\frac{3 x+1}{x+1}=y$
then $y+\frac{1}{y}=\frac{5}{2}$
$\Rightarrow \frac{y^2+1}{y}=\frac{5}{2} $
$ \Rightarrow 2 y^2+2=5 y$
$ \Rightarrow 2 y^2+5 y+2=0 $
$ \Rightarrow 2 y^2-4 y-y+2=0$
$ \Rightarrow 2 y(y-2)-1(y-2)=0 $
$ \Rightarrow(y-2)(2 y-1) 0$
If $y - 2 = 0$ or $2y - 1 = 0$
then $y =2$ or $y=\frac{1}{2}$
$\Rightarrow \frac{3 x+1}{x+1}=2$ or $\frac{3 x+1}{x+1}=\frac{1}{2}$
$\Rightarrow 3 x+1=2 x+2$ or $6 x+2=x+1$
$\Rightarrow x=1$ or $5 x=-1$
$\Rightarrow x=1$ or $x=\frac{-1}{5}$
then $y+\frac{1}{y}=\frac{5}{2}$
$\Rightarrow \frac{y^2+1}{y}=\frac{5}{2} $
$ \Rightarrow 2 y^2+2=5 y$
$ \Rightarrow 2 y^2+5 y+2=0 $
$ \Rightarrow 2 y^2-4 y-y+2=0$
$ \Rightarrow 2 y(y-2)-1(y-2)=0 $
$ \Rightarrow(y-2)(2 y-1) 0$
If $y - 2 = 0$ or $2y - 1 = 0$
then $y =2$ or $y=\frac{1}{2}$
$\Rightarrow \frac{3 x+1}{x+1}=2$ or $\frac{3 x+1}{x+1}=\frac{1}{2}$
$\Rightarrow 3 x+1=2 x+2$ or $6 x+2=x+1$
$\Rightarrow x=1$ or $5 x=-1$
$\Rightarrow x=1$ or $x=\frac{-1}{5}$
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