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Section 2.11 : Linear Inequalities

2. Solve the following inequality and give the solution in both inequality and interval notation.

\[\frac{1}{2}\left( {3 + 4t} \right) \le 6\left( {\frac{1}{3} - \frac{1}{2}t} \right) - \frac{1}{4}\left( {2 + 10t} \right)\]

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Hint : Remember that solving linear inequalities is pretty much the same as solving a linear equation. Just remember to be careful when multiplying/dividing by a negative number.
Start Solution

We know that the process of solving a linear inequality is pretty much the same process as solving a linear equation. We do basic algebraic manipulations to get all the \(t\)’s on one side of the inequality and the numbers on the other side. Just remember that what you do to one side of the inequality you have to do to the other side as well. So, let’s go through the solution process for this linear inequality.

First, we should clear out the parenthesis on both sides and do any simplification that we can. Doing this gives,

\[\begin{array}{c}\displaystyle \frac{3}{2} + 2t \le 2 - 3t - \frac{1}{2} - \frac{5}{2}t\\\displaystyle \frac{3}{2} + 2t \le \frac{3}{2} - \frac{{11}}{2}t\end{array}\] Show Step 2

We can now add \(\frac{{11}}{2}t\) to both sides and subtract \(\frac{3}{2}\) from both sides to get,

\[\frac{{15}}{2}t \le 0\] Show Step 3

For the final step we need to multiply both sides by \(\frac{2}{{15}}\) to get,

\[t \le 0\]

So, the inequality form of the solution is \(\require{bbox} \bbox[2pt,border:1px solid black]{{t \le 0}}\) and the interval notation form of the solution is \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( { - \infty ,0} \right]}}\) .

Remember that we use a square bracket, i.e. “]“, for the right portion of the interval because we are including zero in the solution. Also recall that infinities never get square brackets!