College Algebra (Abridged)

It should be noted that in this case, the sign of inequality has been reversed from ‘less than or equal to’ to ‘greater than or equal to’. Now, let us practice some examples to solve the inequalities. Inequalities Solution + ( + 2) < 16 2 + 2 < 16 Step 1: Simplify the inequality + + 2 < 16 2 < 14 Step 2: Subtract 2 from both sides 2 + 2– 2 < 16– 2 < 7 Step 3: Divide both sides by 2 2 2 < 14 2 5 ≤ 25 ≤ 5 Step 1: Divide both sides by 5 5 5 ≤ 25 5 – 5 ≥ 17 ≥ 22 Step 1: Add 5 to both sides – 5 + 5 ≥ 17 + 5 − 2 + 10 < 64 − 2 < 54 Step 1: Subtract 10 from both sides − 2 + 10– 10 < 64– 10 > − 27 Step 2: Divide both sides by ( − 2 ) & reverse the inequality sign − 2 − 2 > 54 − 2 1.7 Solving Inequalities with Modulus Symbol Solving inequalities involving absolute values needs a little more attention. For instance, 3 > | | means that is 3 units from 0, which means could be 3 or − 3 . The range of values would be > − 3 and < 3 . This can also be written as − 3 < < 3 . This is what is shown on the number line below: Let us look at another example. Solve: |2 − 7| ≥ 5 . 1. Set up a compound inequality: − 5 ≥ 2 − 7 ≥ 5 2. Add seven to both sides: 2 ≥ 2 ≥ 12 3. Divide both sides by 2: 1 ≥ ≥ 6

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