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Sagot :
To solve that inequality we must divide it into two parts, remember that
[tex]|x|=\begin{cases}x,x>0 \\ -x,x<0\end{cases}[/tex]Then we can write
[tex]\begin{gathered} |2x+1|<5\Rightarrow\begin{cases}2x+1<5 \\ -(2x+1)<5\end{cases} \\ \end{gathered}[/tex]Now we have two inequalities:
[tex]\begin{gathered} 2x+1<5 \\ -2x-1<5 \end{gathered}[/tex]Let's solve the first one:
[tex]\begin{gathered} 2x+1<5 \\ \\ 2x<5-1 \\ \\ 2x<4 \\ \\ x<\frac{4}{2} \\ \\ x<2 \end{gathered}[/tex]And the second one
[tex]\begin{gathered} -2x-1<5 \\ \\ -2x<5+1 \\ \\ -2x<6 \\ \\ -x<\frac{6}{2} \\ \\ -x<3 \\ \\ x>-3 \end{gathered}[/tex]Then we have two solutions:
[tex]x<2\text{ and }x>-3[/tex]Writing it in interval notation
[tex]-3
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