The expression given is,
[tex]5|x-3|+3>7[/tex]
Subtract 3 from both sides
[tex]\begin{gathered} 5|x-3|+3-3>7-3 \\ 5|x-3|>4 \end{gathered}[/tex]
Divide both sides by 5
[tex]\begin{gathered} \frac{5|x-3|}{5}>\frac{4}{5} \\ |x-3|>\frac{4}{5} \end{gathered}[/tex]
Apply absolute rule:
[tex]\begin{gathered} x-3<-\frac{4}{5}\text{ or x-3>}\frac{4}{5} \\ \end{gathered}[/tex]
Add 3 to both sides
[tex]\begin{gathered} x-3+3<-\frac{4}{5}+3\text{ or x-3+3>}\frac{4}{5}+3 \\ x<\frac{11}{5}\text{ or x>}\frac{19}{5} \end{gathered}[/tex]
Therefore, the answer has the form:
[tex](-\infty,A)\cup(B,\infty)[/tex]
Hence, the solution using interval notation is
[tex](-\infty,\frac{11}{5})\cup(\frac{19}{5},\infty)[/tex]
