Section 4.2 Compound Inequalities
Inequalities are similar to equations, but involve the relations \(\lt,\leq,\gt,\geq\) instead of equality. We can manipulate these like we do equations, but care must be taken to consider if the inequality changes with the operation performed.
Example 4.7.
Consider the linear inequality
\begin{equation*}
-2x - 4 \lt 8.
\end{equation*}
Adding 4 to both sides preserves the relationship, so that
\begin{equation*}
-2x \lt 12.
\end{equation*}
Dividing by 2 also maintains the inequality
\begin{equation*}
-x \lt 6.
\end{equation*}
However, multiplying both sides by \(-1\) reverses the relation
\begin{equation*}
x \gt - 6.
\end{equation*}
The solution to the inequality is the interval
\((-6,+\infty)\text{.}\)
Example 4.8.
Solve the linear inequality
\begin{equation*}
4-3(x+2) \geq 7
\end{equation*}
Solution.
\begin{align*}
4-3(x+2) \,\amp\geq\, 7\\
-3(x+2) \,\amp\geq\, 3\\
x+2 \,\amp{\color{blue} \leq}\, -1\\
x \,\amp\leq\, -3
\end{align*}
So the solutions are in the interval
\((-\infty,-3]\text{.}\)
A compound inequality involves multiples inequalities joined with the logical connections AND or OR. For example,
\begin{equation*}
3x \lt 7 \quad \text{AND} \quad x \geq 5
\end{equation*}
and
\begin{equation*}
3x \lt 7 \quad \text{OR} \quad x \geq 5
\end{equation*}
are examples of compound inequalities.
Example 4.9. Intervals and Compound Inequalities.
Suppose \(a \lt b\) are real numbers. The interval \((a,b)\) is precisely the solutions to the compound inequality
\begin{equation*}
a \lt x \quad \text{AND} \quad x \lt b.
\end{equation*}
Inequalities involving AND are often abbreviated as
\begin{equation*}
a \lt x \lt b.
\end{equation*}
Example 4.10. Compound Inequality with AND.
Solve the compound inequality
\begin{equation*}
3 \leq 4x + 2 \lt 8.
\end{equation*}
Solution.
\begin{gather*}
3 \leq 4x + 2 \lt 8\\
1 \leq 4x \lt 6\\
\frac{1}{4} \leq x \lt \frac{6}{4}\\
\frac{1}{4} \leq x \lt \frac{3}{2}
\end{gather*}
This says
\(1/4 \leq x\) AND
\(x \lt 3/2\text{.}\) Plotting each interval we see
To be in both intervals,
\(x\) must be in their intersection (See
Section 1.11.) So the solution set is
\begin{equation*}
\left[\frac{1}{4},+\infty\right)
\cap
\left(-\infty,\frac{3}{2}\right) =
\left[\frac{1}{4},\frac{3}{2}\right).
\end{equation*}
In mathematical logic, the connective OR is always assumed to be inclusive in that one or the other or both of the inequalities must be satisfied in order to make the statement true.
Example 4.11. Compound Inequality with OR.
Solve the compound inequality
\begin{equation*}
3x+9 \gt 0 \quad \text{OR} \quad -x + 5 \leq 10
\end{equation*}
Solution.
Work with each inequality in parallel
\begin{gather*}
3x+9 \gt 0 \quad \text{OR} \quad -x + 5 \leq 10\\
3x \gt -9 \quad \text{OR} \quad -x \leq 2\\
3x \gt -9 \quad \text{OR} \quad -x \leq 5\\
x \gt -3 \quad \text{OR} \quad x {\color{blue}\geq} -5
\end{gather*}
Plot each inequality and interpret the results inclusively.
To satisfy the compound inequality,
\(x\) must be in one, or the other, or both of the intervals. Thus,
\(x\) must be in their union (see
Section 1.11)
\begin{equation*}
(-3,+\infty)\cup [-5,+\infty) = [-5,+\infty).
\end{equation*}
So the solution set is
\([-5,+\infty)\text{.}\)
Example 4.12.
Solve the compound inequality
\begin{equation*}
3x+9 \gt 0 \quad \text{AND} \quad -x + 5 \leq 10
\end{equation*}
Solution.
This is the same as Example
Example 4.11, but with AND instead of OR.
\begin{gather*}
3x+9 \gt 0 \quad \text{AND} \quad -x + 5 \leq 10\\
3x \gt -9 \quad \text{AND} \quad -x \leq 2\\
3x \gt -9 \quad \text{AND} \quad -x \leq 5\\
x \gt -3 \quad \text{AND} \quad x \geq -5
\end{gather*}
To satisfy the compound inequality,
\(x\) must be both of the intervals. Thus,
\(x\) must be in their intersection and the solution set is
\([-5,-3)\text{.}\)
Example 4.13. Not all inequalities have solutions.
Solve
\begin{equation*}
5-x \lt 0 \quad \text{AND} \quad -2x \geq 8.
\end{equation*}
Solution.
We require
\begin{equation*}
5 \lt x \quad \text{AND} \quad x \leq -4.
\end{equation*}
Plotting
where we see their intersection is empty. Thus, there is no solution to the compound inequality.
