Answer
The solution set of the rational equation $x+\frac{1}{x}=2$ is $\left\{ 1 \right\}$.
Work Step by Step
$x+\frac{1}{x}=2$
Note that, $x\ne 0$.
The denominators of the rational expressions are $1$, $x$, and 1.
On solving the LCD is $x$.
Multiply both sides by $x$.
$x\left( x+\frac{1}{x} \right)=x\left( 2 \right)$
Apply the distributive property,
$x\cdot \left( x \right)+x\cdot \left( \frac{1}{x} \right)=x\cdot \left( 2 \right)$
Simplify the terms,
$\begin{align}
& {{x}^{2}}+\frac{x}{x}=2x \\
& {{x}^{2}}+1=2x
\end{align}$
Apply the addition principle: Subtracting $2x$ from both the sides of the equation.
$\begin{align}
& {{x}^{2}}+1-2x=2x-2x \\
& {{x}^{2}}-2x+1=0
\end{align}$
Thus, the quadratic equation is formed with 0 on one side.
${{x}^{2}}-2x+1=0$
Factoring the quadratic equation,
$\left( x-1 \right)\left( x-1 \right)=0$
Here, two factors are the same.
Apply the principle of zero products,
$\begin{align}
& \left( x-1 \right)=0\, \\
& x=1
\end{align}$
Now, check the solution of the equation $x+\frac{1}{x}=2$,
Substitute $x=1$ in $x+\frac{1}{x}=2$:
$\begin{align}
1+\frac{1}{1}\overset{?}{\mathop{=}}\,2 & \\
1+1\overset{?}{\mathop{=}}\,2 & \\
2=2 & \\
\end{align}$
So, $x=1$ is the solution of $x+\frac{1}{x}=2$.
