Respuesta :
To answer this question we will use the following property:
[tex]|a|>b>0\text{ if and only if }a>b\text{ or }a<-b.[/tex]Subtracting 6 from the given inequality we get:
[tex]\begin{gathered} 2|4t-1|+6-6>20-6, \\ 2|4t-1|>14. \end{gathered}[/tex]Dividing the above inequality by 2 we get:
[tex]\begin{gathered} \frac{2|4t-1|}{2}>\frac{14}{2}, \\ |4t-1|>7. \end{gathered}[/tex]Then:
[tex]4t-1>7\text{ or }4t-1<-7.[/tex]Solving the above inequalities we get:
1)
[tex]4t-1>7.[/tex]Adding 1 to the above inequality we get:
[tex]\begin{gathered} 4t-1+1>7+1, \\ 4t>8. \end{gathered}[/tex]Dividing the above by 4 we get:
[tex]\begin{gathered} \frac{4t}{4}>\frac{8}{4}, \\ t>2. \end{gathered}[/tex]The above inequality in interval notation is:
[tex](2,\infty).[/tex]2)
[tex]4t-1<-7.[/tex]Adding 1 to the above inequality we get:
[tex]\begin{gathered} 4t-1+1<-7+1, \\ 4t<-6. \end{gathered}[/tex]Dividing the above result by 4 we get:
[tex]\begin{gathered} \frac{4t}{4}<-\frac{6}{4}, \\ t<-\frac{3}{2}. \end{gathered}[/tex]The above inequality in interval notation is:
[tex](-\infty,-\frac{3}{2}).[/tex]Answer:
[tex](-\infty,-\frac{3}{2})\cup(2,\infty).[/tex]