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Outside, with a Fraction

Part of the Inequalities & Absolute Value section of Coddy's Algebra journey. Lesson 60 of 61.

Take ∣x+1∣2≥1\frac{\left|x+1\right|}{2} \ge 1. The absolute value is divided by 2, so multiply both sides by 2 first to get ∣x+1∣≥2\left|x+1\right| \ge 2. The sign is "greater than or equal to", so there are two cases.

x+1≥2x+1 \ge 2

x≥1x \ge 1

or

x+1≤−2x+1 \le -2

x≤−3x \le -3

The solution is x≤−3x \le -3 or x≥1x \ge 1.

Get the bars alone, split into two cases, solve each case, and keep the word "or" between them. A fraction outside the bars is one more step to do. Remove it first, before the split, by multiplying both sides.

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Challenge

Hard

Solve the inequality ∣x−2∣3>1\frac{\left|x-2\right|}{3} > 1 for xx.

Try it yourself

∣x−2∣3>1\frac{\left|x-2\right|}{3} > 1

Get xx alone on one side

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