Compound Inequalities
A compound inequality combines two inequalities as a union or intersection. A union (\(\cup\)) is the combination of values in two sets. An intersection (\(\cap\)) is the set of overlapping values within two sets.
Example 1: Solve and graph the inequality: \(-5\ \le12x+7\le49\).
\(-12\ \le12x\le42\)
\(-1\le x\le\frac{14}{4}\)
A compound inequality combines two inequalities as a union or intersection. A union (\(\cup\)) is the combination of values in two sets. An intersection (\(\cap\)) is the set of overlapping values within two sets.
Example 1: Solve and graph the inequality: \(-5\ \le12x+7\le49\).
\(-12\ \le12x\le42\)
\(-1\le x\le\frac{14}{4}\)
This solution is an intersection and we only shade the overlap.
Graph:
Graph:
Example 2: Solve and graph the inequality: \(-11x\ge12\ \text{or}\ \ 2x-6\le18\).
\(x\le-\frac{12}{11}\ \text{or}\ \ 2x\le24\)
\(x\le-\frac{12}{11}\ \text{or}\ \ x\le12\)
This solution is a union and so we shade everything that is contained in both solutions.
Graph:
Example 3: Solve and graph the inequality \(12-x\ge10\ \text{and}\ 2x-7\ge3\).
\(-x\ge-2\ \text{and}\ 2x\ge10\)
\(x\le2\ \text{and}\ x\ge5\)
Because this is supposed to be an intersection denoted by the word "and," and there is not overlap for \(x\le2\ \text{and}\ x\ge5\) there is no solution.
Graph: