Given:
[tex]3-2\lvert7x-3\rvert<-1[/tex]
Solution:
First, we need to transfer 3 to the other side.
[tex]\begin{gathered} 3-2\lvert7x-3\rvert<-1 \\ -2\lvert7x-3\rvert<-1-3 \\ -2\lvert7x-3\rvert<-4 \end{gathered}[/tex]
Then, we will divide both sides with -2 to isolate the terms with absolute value.
[tex]\begin{gathered} -2\lvert7x-3\rvert<-4 \\ \frac{-2\lvert7x-3\rvert}{-2}<\frac{-4}{-2} \\ \lvert7x-3\rvert>2 \end{gathered}[/tex]
Now, in dealing with absolute value, we will set it's limits.
[tex]\begin{gathered} \lvert7x-3\rvert>2 \\ \text{absolu}te\text{ value can also be written as:} \\ -2>7x-3>2 \end{gathered}[/tex]
Let's equate the inequalities separately.
[tex]\begin{gathered} 7x-3>2 \\ 7x>2+3 \\ 7x>5 \\ \frac{7x}{7}>\frac{5}{7} \\ x>\frac{5}{7} \end{gathered}[/tex]
For the other equation:
[tex]\begin{gathered} 7x-3<-2 \\ 7x<-2+3 \\ 7x<1 \\ \frac{7x}{7}<\frac{1}{7} \\ x<\frac{1}{7} \end{gathered}[/tex]
Therefore, we will have x is less than 1/7 and greater than 5/7.
ANSWER:
[tex](-\infty,\frac{1}{7})\cup(\frac{5}{7},\infty)[/tex]
