The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 5 (Intermediate Algebra) Solving Inequalities and Absolute Value inequalities

Overview

This lecture extends equation-solving techniques to inequalities. Solving a linear inequality works almost exactly like solving a linear equation, with one crucial difference: multiplying or dividing both sides by a negative number reverses the inequality sign. You will also learn to represent solution sets three ways — on a number line, in interval notation, and as compound inequalities joined by "and" (intersection) or "or" (union). The lecture closes with absolute-value inequalities, which always unpack into a pair of inequalities: a "less than" absolute value becomes an "and" statement, while a "greater than" absolute value becomes an "or" statement.

Key Concepts

Inequality symbols. The four relations are $<$ (less than), $\le$ (less than or equal to), $>$ (greater than), and $\ge$ (greater than or equal to). A statement like $x < 2$ describes all numbers smaller than 2 — infinitely many values, not a single answer.

Graphing on a number line.

• A strict inequality ($<$ or $>$) uses an open circle (or a rounded bracket) at the endpoint — the endpoint is not included.
• A non-strict inequality ($\le$ or $\ge$) uses a filled circle (or a square bracket) — the endpoint is included.
• Shade the side of the line that satisfies the inequality.

Interval notation.

• Parentheses ( ) exclude an endpoint; square brackets [ ] include it.
• Infinity is always written with a parenthesis: $\infty$ and $-\infty$ are never reached.
• Example: $x < 2$ is $(-\infty, 2)$; $x \ge -3$ is $[-3, \infty)$.

Solving linear inequalities. Use the same moves as for equations — add or subtract the same quantity on both sides, multiply or divide both sides by the same number.

• Adding/subtracting any number, or multiplying/dividing by a positive number, leaves the inequality direction unchanged.
• Multiplying or dividing both sides by a negative number flips the inequality sign. This is the single most common mistake to guard against. For example, $-x \ge 2$ becomes $x \le -2$ after multiplying by $-1$.

Compound inequalities ("and" vs. "or").

• "And" (intersection): a solution must satisfy both inequalities. The graph is a single connected segment. Example: $x > -2$ and $x < 9$ gives $(-2, 9)$.
• "Or" (union): a solution need satisfy only one inequality. The graph splits into two separate pieces. Example: $x < -2$ or $x > 9$ gives $(-\infty, -2) \cup (9, \infty)$.

Absolute-value inequalities. Recall that $|x|$ measures distance from zero, so $|x|$ is never negative.

• "Less than" ⟹ "and": $|x| \le a$ (with $a \ge 0$) means $-a \le x \le a$. In general, $|\text{expr}| \le a$ becomes $-a \le \text{expr} \le a$.
• "Greater than" ⟹ "or": $|x| > a$ means $x > a$ or $x < -a$. In general, $|\text{expr}| > a$ becomes $\text{expr} > a$ or $\text{expr} < -a$.
• No-solution case: an absolute value can never be less than a negative number, so $|\text{expr}| < (\text{negative})$ has no solution (the empty set $\varnothing$). Always isolate the absolute value first before splitting.

Worked Example

Solve the "or" inequality $2x - 5 > 3$ or $4 - x \ge 6$.

Solve each piece separately, remembering to flip the sign when dividing by a negative.

Because the pieces are joined by "or," take the union of the two solution sets:

$x \le -2 \;\text{ or }\; x > 4 \quad\Longrightarrow\quad (-\infty,\, -2] \,\cup\, (4,\, \infty)$

Absolute-value example: solve $|x + 1| < 7$. Since this is a "less than" absolute value, rewrite it as a single "and" (a three-part inequality) and solve:

The solution is $(-8,\, 6)$.

Summary

• Solve inequalities exactly like equations, but flip the inequality sign whenever you multiply or divide both sides by a negative number.
• Open circles / parentheses exclude an endpoint; filled circles / square brackets include it; $\infty$ always takes a parenthesis.
• An "and" compound inequality is an intersection — one connected segment; an "or" is a union — two separate pieces.
• $|\text{expr}| \le a$ unpacks into the "and" statement $-a \le \text{expr} \le a$.
• $|\text{expr}| > a$ unpacks into the "or" statement $\text{expr} > a$ or $\text{expr} < -a$.
• Always isolate the absolute value first; if it must be less than a negative number, there is no solution (the empty set).