Solving and Modeling Linear Inequalities — SAT Math

Introduction

Linear inequalities appear throughout the SAT's Algebra domain (roughly 35% of questions). Every real-world constraint — a spending cap, a weight limit, a minimum score — eventually becomes an inequality. This note covers three tightly linked skills: solving inequalities algebraically (including the sign-flip rule), translating English phrases into inequality symbols, and interpreting solutions in context, particularly when only whole-number answers make sense. For graphing inequality regions in the $xy$ -plane and working with systems of inequalities, see the sibling note Graphs of Linear Inequalities and Systems.

Core Concept

Solving a one-variable inequality

Solving a linear inequality works exactly like solving a linear equation (see Solving Linear Equations) with one critical difference:

When you multiply or divide both sides by a negative number, reverse the inequality sign.

Why? Multiplying by $-1$ reflects the number line, swapping the left-right order of every value.

Quick illustration:

-3x > 9

Divide both sides by $-3$ and flip $>$ to $<$ :

x < -3

Check: $x = -4 \Rightarrow -3(-4) = 12 > 9$ ✓ $\quad x = -2 \Rightarrow -3(-2) = 6 < 9$ ✗

Translating English to symbols

English phrase	Symbol
at least $k$	$\geq k$
at most $k$	$\leq k$
no more than $k$	$\leq k$
more than $k$	$> k$
fewer than / less than $k$	$< k$
between $a$ and $b$ (inclusive)	$a \leq x \leq b$

"At least" and "at most" always use non-strict inequalities ( $\leq$ or $\geq$ ) because they include the boundary value.

Two-variable inequalities as constraints

When two quantities are linked — say, $x$ items at $8 each and $y$ items at $5 each — a budget cap of $200 becomes:

8x + 5y \leq 200

Every ordered pair $(x, y)$ satisfying this is a solution. Real-world context often further restricts $x$ and $y$ to non-negative integers.

Whole-number restriction

If the variable counts discrete objects (people, crates, tickets), the mathematical solution $n \leq 7.4$ means the maximum is $n = 7$ — always round inward (down for a maximum, up for a minimum) to stay within the constraint.

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