Question

Solve the equation $x\left(\frac{3-x}{x+1}\right)(x+\frac{3-x}{x+1})=2$

Answer 1

Given equation is defined when $x+1\ne 0$

Let $x\left(\frac{3-x}{x+1}\right)=u$ and $x+\frac{3-x}{x+1}=v$

$\therefore$ the given equation is $uv=2$ ..... (1)

Now, $u+v=x\left(\frac{3-x}{x+1}\right)+x+\left(\frac{3-x}{x+1}\right)$

$=(x+1)\left(\frac{3-x}{x+1}\right)+x$

$=3-x+x=3$
$\therefore u+v=3$ .....(2)

Substituting the value of $u$ from (2) in (1),

$(3-v)v=2$

$\Rightarrow v^2-3v+2=0$

$\Rightarrow (v-1)(v-2)=0$

$\Rightarrow v=1$ or $v=2$

So, $u=2,v=1$ or $u=1,v=2$

Given equation is equivalent to the collection
$\therefore \{\begin{array}{c}x\left(\frac{3-x}{x+1}\right)=2\\ x+\frac{3-x}{x+1}=1\end{array}$ or $\{\begin{array}{c}x\left(\frac{3-x}{x+1}\right)=1\\ x+\frac{3-x}{x+1}=2\end{array}$

$\Rightarrow \{\begin{array}{c}{x}^2-x+2=0\\ x^2-x+2=0\end{array}$ or $\{\begin{array}{c}{x}^2-2x+1=0\\ x^2-2x+1=0\end{array}$

$\Rightarrow \{\begin{array}{c}{x}^2-x+2=0\\ x^2-2x+1=0\end{array}$

$\Rightarrow \{\begin{array}{c}{(x-\frac{1}{2})}^2+\frac{7}{4}=0\\ {(x-1)}^2=0\end{array}$

But ${(x-\frac{1}{2})}^2+\frac{7}{4}\ne 0$
$\therefore {(x-1)}^2=0$

$\Rightarrow x=1$ is a unique solution of the original equation.