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Solve each inequality). 2|4t-1|+6>20

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To answer this question we will use the following property:


|a|>b>0\text{ if and only if }a>b\text{ or }a<-b.

Subtracting 6 from the given inequality we get:


\begin{gathered} 2|4t-1|+6-6>20-6, \\ 2|4t-1|>14. \end{gathered}

Dividing the above inequality by 2 we get:


\begin{gathered} (2|4t-1|)/(2)>(14)/(2), \\ |4t-1|>7. \end{gathered}

Then:


4t-1>7\text{ or }4t-1<-7.

Solving the above inequalities we get:

1)


4t-1>7.

Adding 1 to the above inequality we get:


\begin{gathered} 4t-1+1>7+1, \\ 4t>8. \end{gathered}

Dividing the above by 4 we get:


\begin{gathered} (4t)/(4)>(8)/(4), \\ t>2. \end{gathered}

The above inequality in interval notation is:


(2,\infty).

2)


4t-1<-7.

Adding 1 to the above inequality we get:


\begin{gathered} 4t-1+1<-7+1, \\ 4t<-6. \end{gathered}

Dividing the above result by 4 we get:


\begin{gathered} (4t)/(4)<-(6)/(4), \\ t<-(3)/(2). \end{gathered}

The above inequality in interval notation is:


(-\infty,-(3)/(2)).

Answer:


(-\infty,-(3)/(2))\cup(2,\infty).

User Hivaga
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