Linear Inequalities
Inequalities work like equations — you isolate the variable using the same algebra moves — with one critical exception that trips up almost every student under pressure. This note covers the four flavors of inequality the SAT actually tests: one-variable, two-variable, compound, and systems. It also covers the single rule that, when forgotten, costs the most points.
Single-variable inequalities
An inequality with one variable is solved exactly like its corresponding equation — same algebra moves, same order of operations. The solution is the set of all values that make the inequality true.
Apply the same operations to both sides until the variable stands alone. The solution is read directly from the resulting inequality.
On a number line, the solution shades everything to the left of 4. The circle at 4 is open (not included) because the inequality is strict.
Open vs. filled circle: use an open circle (○) when the inequality is strict (< or >). Use a filled circle (●) when it includes equality (≤ or ≥).
When you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign. This is the single most-missed rule on SAT inequality questions.
This rule does NOT apply when adding or subtracting a negative — only when multiplying or dividing.
The flip rule appears whenever your isolating step requires dividing or multiplying by a negative coefficient. Watch for it whenever the variable starts with a negative coefficient.
- Subtract 7 from both sides:
−3x ≥ 6
- Divide both sides by −3 — flip the sign:
x ≤ −2
Two-variable inequalities
An inequality with two variables, like y > 2x − 1, doesn't have a single solution — it has an entire region of the coordinate plane as its solution. Every (x, y) point in that region satisfies the inequality.
- Graph the boundary line as if it were an equation (here, y = 2x − 1).
- Use a dashed line if the inequality is strict (< or >). Use a solid line if it includes equality (≤ or ≥).
- Shade above the line if y > (something), below if y < (something).
Not sure which side to shade? Plug in the point (0, 0) into the inequality. If it makes a true statement, shade the side that contains the origin. If false, shade the other side.
For y > 2x − 1: (0, 0) gives 0 > −1, which is true → shade the side containing the origin.
Compound inequalities
A compound inequality combines two conditions. There are two types: "and" (intersection — must satisfy both) and "or" (union — must satisfy at least one).
The variable must satisfy BOTH conditions simultaneously. Often written compactly as a single chain:
This means x ≥ −2 AND x < 5. The solution is everything between −2 (included) and 5 (not included).
The variable satisfies AT LEAST ONE condition. Written as two separate inequalities:
The solution covers two disconnected regions of the number line.
For a chain like A < expression < B, apply the same operation to all three parts simultaneously.
- Subtract 2 from each part:
−6 < 3x ≤ 9
- Divide each part by 3:
−2 < x ≤ 3
Systems of inequalities
A system of inequalities is two or more inequalities considered together. The solution is the region where ALL inequalities are satisfied — the intersection of their individual solution regions.
Common SAT setup:
Graph each region separately, then identify the overlap. Points in the overlap satisfy both inequalities.
Inequalities from word problems
The hardest SAT inequality problems aren't algebraic — they're word problems where you have to translate English constraints into algebraic inequalities. The key is recognizing the signal phrases.
| Phrase | Inequality |
|---|---|
| "At most" / "no more than" / "cannot exceed" | ≤ |
| "At least" / "no less than" / "minimum of" | ≥ |
| "Less than" / "fewer than" | < |
| "More than" / "greater than" | > |
| "Between A and B" | A < x < B (check context for inclusive) |
A store sells two types of pens. Type A costs $2 and Type B costs $3. A customer wants to buy at least 10 pens total but spend no more than $24. If a represents the number of Type A pens and b represents Type B pens, write a system of inequalities representing this situation.
- "At least 10 pens" — total of a + b is at least 10:
a + b ≥ 10
- "Spend no more than $24" — total cost is at most $24:
2a + 3b ≤ 24
- Quantities can't be negative:
a ≥ 0, b ≥ 0
The solution is any (a, b) point that satisfies all four inequalities.
Sample SAT-style problems
Which of the following is equivalent to the inequality 5 − 4x ≥ 13?
- (A) x ≥ −2
- (B) x ≤ −2
- (C) x ≥ 2
- (D) x ≤ 2
- Subtract 5 from both sides:
−4x ≥ 8
- Divide by −4 — flip the sign:
x ≤ −2
Which point is a solution to the inequality 3x − 2y > 6?
- (A) (0, 0)
- (B) (1, −2)
- (C) (2, 0)
- (D) (4, 4)
- Test (A): 3(0) − 2(0) = 0. Is 0 > 6? No.
- Test (B): 3(1) − 2(−2) = 3 + 4 = 7. Is 7 > 6? Yes ✓
A delivery van's payload cannot exceed 1,500 pounds. The van is currently carrying 200 pounds of equipment. If each box weighs 35 pounds, what is the maximum number of boxes b the van can hold without exceeding the payload limit?
- Set up the inequality. Total weight = 200 + 35b, must be ≤ 1500:
200 + 35b ≤ 1500
- Solve:
35b ≤ 1300 b ≤ 37.14...
- Since b must be a whole number of boxes, take the floor.
Is the point (1, 2) in the solution region of the system below?
- Test the first inequality: 2 > 2(1) − 3 = −1. ✓
- Test the second inequality: 2 ≤ −(1) + 4 = 3. ✓
- Both inequalities are satisfied.
1. Forgetting to flip the sign when multiplying or dividing by a negative. Train yourself to physically circle the negative coefficient before solving — it'll catch most of these.
2. Misreading "at most" / "at least" on word problems. "At most 10" means ≤ 10 (10 is allowed). "Fewer than 10" means < 10 (10 is not allowed). The SAT specifically writes wrong-answer choices that swap these.