In equation involving absolute value one solution might get rejected, why is that? For example the equation $|2x-1|+3=4x$,
We can get two equations out of that, $2x-1=4x-3$ and $2x-1=-(4x-3)$. Solve the first one we get $x=1$ and for the second one $x=2/3$. But when you substitute 2/3 into the original equation, you get 10/3=8/3, which is clearly wrong. My answer sheet says the answer is rejected, but if we follow all steps correctly and get to a solution, how can it still be wrong? It's like solving $5=3+x$ and x somehow doesn't equal to two.
 A: Because:
$$|2x-1|+3=4x \implies 2x-1=4x-3 \lor 2x-1=-(4x-3)$$
and the converse is false.
Which means that any solution of LHS is also solution of RHS, but the converse is not necessary true. This is why if you perform a non-reversible derivation, you need to check the solutions in the original equation.
A: The absolute value of a number cannot be negative.
Observe that $2x - 1 \geq 0 \implies x \geq \dfrac{1}{2}$.
Hence, $|2x - 1| = 2x - 1$ if $x \geq 1/2$.  On the other hand, if $x < 1/2$, then $2x - 1 < 0$, so $|2x - 1| = -(2x - 1) = -2x + 1 > 0$.  Therefore, we obtain the piecewise definition
$$
|2x - 1| = \begin{cases}
           2x - 1 & \text{if $x \geq \dfrac{1}{2}$}\\[2 mm]
           -2x + 1 & \text{if $x < \dfrac{1}{2}$}
           \end{cases}
$$
Thus, we must consider cases, depending on whether $x \geq 1/2$.
Case 1: If $x \geq 1/2$,
\begin{align*}
|2x - 1| + 3 & = 4x\\
|2x - 1| & = 4x - 3\\
2x - 1 & = 4x - 3\\
2 & = 2x\\
1 & = x
\end{align*}
Since $1 \geq 1/2$, $x = 1$ is a solution, as you can verify by direct substitution.
Case 2: If $x < 1/2$,
\begin{align*}
|2x - 1| + 3 & = 4x\\
|2x - 1| & = 4x - 3\\
-2x + 1 & = 4x - 3\\
4 & = 6x\\
\frac{2}{3} & = x
\end{align*}
However, $2/3 \geq 1/2$, so it is not satisfy the restriction that $x < 1/2$.  Therefore, $x = 2/3$ is not a valid solution, as you can verify by direct substitution.
