Exercise 24
The solution set of the inequation \( \left| \cfrac{2x + 1}{3} \right| < 5 \) is the following:
The solution set of the inequation:
\[
\left| \cfrac{2x + 1}{3} \right| < 5,
\]
is the following:
-
\( ( -\infty, \, 7 ) \)
-
\( ( -8, +\infty ) \)
-
\( ( -\infty, -8 ) \)
-
\( ( -7, \, 8 ) \)
-
\( ( -8, \, 7 ) \)
Try to solve it before checking the answer.
-
\( ( -8, \, 7 ) \)
Indeed:
\[
\begin{aligned}
\left| \frac{2x + 1}{3} \right| < 5
&\Leftrightarrow
-5 < \frac{2x + 1}{3} < 5
\\
&\hspace{1em}\text{Inequalities}
\\[1.5em]
&\Leftrightarrow
-15 < 2x + 1 < 15
\\
&\hspace{1em}\text{multiplying by 3}
\\[1.5em]
&\Leftrightarrow
-16 < 2x < 14
\\
&\hspace{1em}\text{adding -1}
\\[1.5em]
&\Leftrightarrow
-8 < x < 7
\\
&\hspace{1em}\text{dividing by 2}
\\[1.5em]
&\Leftrightarrow
\boldsymbol{
x \in (-8, \, 7)
}
\end{aligned}
\]
\[
\begin{aligned}
\left| \frac{2x + 1}{3} \right| < 5
&\Leftrightarrow
-5 < \frac{2x + 1}{3} < 5
& \text{Inequalities}
\\[.5em]
&\Leftrightarrow
-15 < 2x + 1 < 15
&\text{multiplying by 3}
\\[.5em]
&\Leftrightarrow
-16 < 2x < 14
&\text{adding -1}
\\[.5em]
&\Leftrightarrow
-8 < x < 7
&\text{dividing by 2}
\\[.5em]
&\Leftrightarrow
\boldsymbol{
x \in (-8, \, 7)
}
\end{aligned}
\]