###### Example3.5.3

Solve for \(x\) in \(2x+1=2x+1\text{.}\)

We will move all terms containing \(x\) to one side of the equals sign:

\begin{align*}
2x+1\amp=2x+1\\
2x+1\subtractright{2x}\amp=2x+1\subtractright{2x}\\
1\amp=1
\end{align*}

At this point, \(x\) is no longer contained in the equation. What value can we substitute into \(x\) to make \(1=1\) true? Any number! This means that all real numbers are possible solutions to the equation \(2x+1=2x+1\text{.}\) We say this equation's solution set contains *all real numbers*. We can write this set using set-builder notation as \(\{x\mid x\text{ is a real number}\}\) or using interval notation as \((-\infty,\infty)\text{.}\)

The equation \(1=1\) is known as an identity as it is always true.