Linear Inequalities

Almost the same — but one law bites

From the Order and Inequality lesson in Bedrock, here is the key theorem: multiplying both sides of an inequality by a negative number reverses its direction. We proved it there with the number-line flip. Make it concrete:

$3 < 7. \quad \text{Multiply both sides by }-1: \quad -3 \quad ? \quad -7.$

Is $-3 < -7$? No — $-3 > -7$. The order flipped. This isn't invented; it's the number line announcing that negation reverses left and right.

The one new law: when you multiply or divide both sides of an inequality by a negative number, flip the inequality sign. Everything else is identical to solving equations.

Solving and expressing solution sets

Solve $3x - 7 < 5$:

$3x < 12 \implies x < 4.$

The solution is not one number — it's an infinite set. Interval notation: $(-\infty, 4)$. On the number line: open circle at 4 (4 is not a solution since $3(4)-7 = 5 \not< 5$), shading to the left.

Now solve $-2x + 3 \le 9$:

$-2x \le 6 \implies x \ge -3.$

Dividing by $-2$ (negative) flips $\le$ to $\ge$. Interval: $[-3, +\infty)$. Closed circle at $-3$ because $-2(-3)+3 = 9 \le 9$ — yes, $-3$ is a solution.

Drag the point and watch it fall inside or outside the solution region. Open endpoints match strict inequalities; closed endpoints match non-strict ones.

Compound inequalities: intersection and union

"And" inequalities require both conditions simultaneously — this is the intersection of two sets, exactly the $\cap$ operation from the Set Operations lesson.

Solve $-3 < 2x + 1 \le 7$ — work on all three parts at once:

$-4 < 2x \le 6 \implies -2 < x \le 3.$

Interval: $(-2, 3]$. The intersection of the ray $x > -2$ and the ray $x \le 3$.

"Or" inequalities require at least one condition — this is the union $\cup$.

Solve "$x - 1 < -3$ or $x + 2 > 6$":

Left piece: $x < -2$. Right piece: $x > 4$. Union: $(-\infty, -2) \cup (4, +\infty)$.

Absolute-value inequalities

The Absolute Value and Distance lesson gave you the key translation: $|x - a|$ is the distance from $x$ to $a$. Use it here.

Band type ($<$): $|x - a| < r$ means within distance $r$ of $a$: $a - r < x < a + r.$

Two-rays type ($>$): $|x - a| > r$ means farther than $r$ from $a$: $x < a - r \quad \text{or} \quad x > a + r.$

Example: Solve $|2x - 3| \le 7$: $-7 \le 2x - 3 \le 7 \implies -4 \le 2x \le 10 \implies -2 \le x \le 5.$

Interval: $[-2, 5]$. Check: $x = -2$: $|2(-2)-3| = |-7| = 7 \le 7$. Yes. $x = 5$: $|2(5)-3| = 7 \le 7$. Yes. $x = 0$: $|-3| = 3 \le 7$. Yes.

Always translate to distance language first. The algebra follows mechanically.

Writing solution sets: interval notation and set-builder

Two equivalent notations for the same infinite set of solutions:

Interval notation. Four building blocks:

• $(a, b)$: all $x$ with $a < x < b$ (both endpoints excluded)
• $[a, b]$: all $x$ with $a \le x \le b$ (both endpoints included)
• $(-\infty, a)$: all $x < a$; $(a, +\infty)$: all $x > a$ (infinity is never included)
• Unions: $(-\infty, a) \cup (b, +\infty)$ for "x < a or x > b"

The relationship to the real line. Every bounded interval is a connected piece of $\mathbb{R}$. Every unbounded ray is a half-line. These are not ad hoc notations — they are the standard descriptions of connected subsets of $\mathbb{R}$, the same sets you'll use when specifying domains of functions, domains of convergence of series, and much more later.

Why this matters more than it looks. The solution set of an inequality is a SET — and everything you know about sets from the Sets stratum applies. "And" means intersection, "or" means union, "not" would mean complement. Set operations are not just set theory homework; they are the operating manual for combining conditions. When you later solve systems of inequalities, you will be computing intersections of half-planes in $\mathbb{R}^2$ — the exact same move, one dimension higher.