Section 2.11 : Linear Inequalities
1. Solve the following inequality and give the solution in both inequality and interval notation.
\[4\left( {z + 2} \right) - 1 > 5 - 7\left( {4 - z} \right)\]Show All Steps Hide All Steps
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 \(z\)’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,
\[\begin{array}{c}4z + 8 - 1 > 5 - 28 + 7z\\ 4z + 7 > - 23 + 7z\end{array}\] Show Step 2We can now subtract 7\(z\) from both sides and subtract 7 to both sides to get,
\[ - 3z > - 30\]Note that we could just have easily subtracted 4\(z\) from both sides and added 23 to both sides. Each will get the same result in the end.
Show Step 3For the final step we need to divide both sides by -3. Recall however that because we are dividing by a negative number we need to switch the direction of the inequality to get,
\[z < 10\]So, the inequality form of the solution is \(\require{bbox} \bbox[2pt,border:1px solid black]{{z < 10}}\) and the interval notation form of the solution is \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( { - \infty ,10} \right)}}\) .
Remember that we use a parenthesis, i.e. “)“, for the right side of the interval notation because we are not including 10 in the solution. Also recall that infinities always get parenthesis!