Split the compound inequality into two separate ones and solve them individually.
Solution: 14 ≤ k ≤ 26 Graph:
Practice makes perfect
We are given the following compound inequality.
15 ≤ 20+11+k/3≤ 19
Sometimes, it can be helpful to write a compound inequality as two individual inequalities. This lets us solve each one separately.
15 ≤ 20+11+k/3 and 20+11+k/3 ≤ 19Now we can solve them separately, and later combine their solution sets.
First Inequality
Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign.
The second inequality is true for numbers less than or equal to 26. Notice that k can equal 26 as the inequality is non-strict.
Combining Solution Sets
Finally, we can combine the obtained solution sets. The first inequality, k ≥ 14, describes all values greater than or equal to 14. To show this on a number line, we can mark a closed circle at 14 —because k can equal 14 — and shade the region to the right.
The second inequality, k ≤ 26, describes all values less than or equal to 26. To show this on a number line, we can mark a closed circle at 26 — because k can equal 26 — and shade the region to the left.
The intersection of these solution sets, 14 ≤ k ≤ 26, describes the solution set to the compound inequality.
