Linear Inequalities

Inequalities Are Equations with a Range of Answers

A linear inequality looks almost exactly like a linear equation, except the equals sign is replaced by $<$, $>$, $\le$, or $\ge$. Instead of a single solution, an inequality describes a whole range of values. On the SAT, inequalities appear in one variable (solve for a range on a number line), in two variables (shade a region in the $xy$-plane), and inside word problems that use phrases like at least or no more than. All of them reward the same careful algebra.

Solving One-Variable Inequalities

You solve a linear inequality using the same moves as an equation — add, subtract, multiply, divide — with one crucial exception: when you multiply or divide both sides by a negative number, you must flip the inequality sign. This is the rule the SAT tests most often, and forgetting it is the single most common error. Note the trap carefully: you flip only for multiplication or division by a negative. Adding or subtracting a negative number never changes the direction.

Worked example: solve $-3x + 7 \ge 22$.

Step 1 — Subtract 7 from both sides: $-3x \ge 15$

Step 2 — Divide both sides by $-3$, and flip the sign because we divided by a negative: $x \le -5$

Step 3 — Interpret. The solution is every value of $x$ that is $-5$ or smaller. Check with a test value like $x = -6$: $-3(-6) + 7 = 18 + 7 = 25 \ge 22$. True, so the direction is correct.

Whenever you finish, plug in a number from your answer range to confirm the direction — it takes five seconds and catches the flip error instantly.

Graphing on a Number Line: Open vs. Closed

When you graph a one-variable inequality, the type of endpoint matters:

Symbol	Endpoint	Meaning
$<$ or $>$	Open circle	Endpoint not included
$\le$ or $\ge$	Closed (filled) circle	Endpoint is included

For $x \le -5$, you place a closed dot at $-5$ and shade everything to the left. For $x > 2$, you place an open dot at $2$ and shade to the right.

Compound Inequalities

A compound inequality such as $-1 \le 2x - 5 < 7$ is really two inequalities at once, and you keep everything balanced by doing the same operation to all three parts. Add 5 across the board to get $4 \le 2x < 12$, then divide every part by 2 to get $2 \le x < 6$. The solution is the interval between two numbers — closed on the left (from $\le$) and open on the right (from $<$).

Two-Variable Inequalities and Shading

A two-variable inequality such as $y > 2x - 1$ describes a half-plane. To graph it:

1. Graph the boundary line $y = 2x - 1$. Use a dashed line for strict inequalities ($<$, $>$) and a solid line for inclusive ones ($\le$, $\ge$).
2. Pick a test point not on the line — $(0, 0)$ is easiest when the line does not pass through the origin.
3. If the test point makes the inequality true, shade the side containing it; if false, shade the other side.

Worked example: which region satisfies $y \ge -x + 4$? Test $(0, 0)$: is $0 \ge -0 + 4$, i.e. $0 \ge 4$? That is false, so $(0, 0)$ is not in the solution. Shade the side of the solid line away from the origin.

Systems of Inequalities

The SAT also asks which point is (or is not) a solution to a system of inequalities. For those, simply substitute the point into every inequality; it is a solution only if all of them are true at once. A point that satisfies just one inequality does not count.

Translating Word Problems

The last skill is turning English into an inequality. Learn these signal phrases: at least and no less than mean $\ge$; at most and no more than mean $\le$; more than means $>$; fewer than / less than means $<$. A sentence like "the total must be at least 90 points" becomes $\text{total} \ge 90$, and from there it is ordinary algebra.

Common Mistakes to Avoid

• Forgetting to flip the sign after multiplying or dividing by a negative number.
• Flipping the sign when you only added or subtracted a negative — that never changes the direction.
• Using a filled circle (or solid line) for a strict $<$ or $>$ inequality — those get an open circle or dashed line.
• Assuming a point solves a system when it satisfies only one of the inequalities.