Respuesta :
From the definition of absolute value, we have the following:
[tex]\begin{gathered} |a|=\mleft\{\begin{aligned}a\text{ if a}\ge0 \\ \square \\ -a\text{ if a}\leq0\end{aligned}\mright. \\ \end{gathered}[/tex]For this case we have:
[tex]|2x-1|=\mleft\{\begin{aligned}2x-1\text{ if 2x-1}\ge0 \\ -(2x-1)\text{ if 2x-1}\leq0\end{aligned}\mright.[/tex]Therefore, we must solve for both cases. First, when 2x-1 is greater or equal to 0, we have:
[tex]\begin{gathered} 2x-1-3\leq-3 \\ \Rightarrow2x-4\leq-3 \\ \Rightarrow2x\leq-3+4=1 \\ \Rightarrow2x\leq1\Rightarrow x\leq\frac{1}{2} \end{gathered}[/tex]Now, for the second case we get:
[tex]\begin{gathered} -(2x-1)-3\leq-3 \\ \Rightarrow-2x+1-3\leq-3 \\ \Rightarrow-2x-2\leq-3 \\ \Rightarrow-2x\leq-3+2=-1 \\ \Rightarrow x\ge-\frac{1}{-2}=\frac{1}{2} \end{gathered}[/tex]Notice how in this case we had to change the orientation of the sign since we divided by -2.
Since we have that x <= 1/2 and x>=1/2, then x=1/2
