title: “Absolute‑Value Inequality” description: “AMC 10 Algebra: Absolute‑Value Inequality” date: 2024-01-01 draft: false type: “notes” categories: [“math”, “amc”, “algebra”] tags: [“amc10”, “algebra”, “absolute-value”, “inequality”, “mathematics”] weight: 1

Absolute‑Value Inequality

1. Identify the critical point(s)

The expression inside the absolute value, $3x-7$, changes sign at

$$ 3x-7=0 ;\Longrightarrow; x=\frac73. $$

2. Split into cases at the breakpoint $x=\dfrac73$

Case A: $x\ge \dfrac73$

Here $3x-7\ge0$, so $|3x-7|=3x-7$.

Inequality becomes

$$ 3x-7 ;\le; 2x+5 ;\Longrightarrow; x\le12. $$

Within this case the variable is constrained simultaneously by

$$ \frac73 ;\le; x;\le; 12. $$

Case B: $x<\dfrac73$

Now $3x-7<0$, so $|3x-7|=-(3x-7)=7-3x$.

Inequality becomes

$$ 7-3x ;\le; 2x+5 ;\Longrightarrow; 5x;\ge;2 ;\Longrightarrow; x;\ge;\frac25. $$

Combine with the case condition to get

$$ \frac25 ;\le; x ;<;\frac73. $$

3. Consolidate the solution

Both cases together give the single interval

$$ \boxed{\dfrac25 ;\le; x ;\le; 12}. $$

(The “case‑split at the sign‑change point” avoided squaring or expanding anything; we just treated each linear piece separately.)

Why this method is fast

• One breakpoint $x=\tfrac73$ produced exactly two linear inequalities, each solved in one step.
• We never squared the inequality (which could introduce extraneous solutions) or manipulated nested absolute values.

Whenever you see an absolute‑value equation or inequality, find the zeroes of the inner expressions first, split the real line at those points, and work piece‑by‑piece. That’s the standard < 60‑second AMC trick.