Linear inequalities in one or two variables
Algebra is about 35% of SAT Math (13–15 of 44 questions). Linear inequalities appear in both modules, often as a word problem about a budget or a limit, or as a question about which point lies in the solution region.
Words and symbols
The SAT hides the symbol inside words. Learn this list. “At least” means the value can be that number or bigger, so ≥. “At most” means that number or smaller, so ≤. “More than” and “fewer than” do not allow the number itself, so they are > and <. In stories about money, a limit such as “can spend no more than $50” means cost ≤ 50.
| Words | Symbol |
|---|---|
| at least, minimum, no less than | ≥ |
| at most, maximum, no more than, up to | ≤ |
| more than, greater than, exceeds | > |
| fewer than, less than, under | < |
Solve it like an equation, with one new rule
Add, subtract, multiply and divide on both sides, as in an equation. There is one new rule: when you multiply or divide both sides by a negative number, the inequality sign turns around. Example: −3x + 5 > 17 gives −3x > 12; dividing by −3 gives x < −4. Check with a number: x = −5 gives −3(−5) + 5 = 20, and 20 > 17 ✓. Adding or subtracting a negative number does not flip the sign.
Stories: set up, solve, round the right way
Write the total in the story as an expression, put the correct symbol, then solve. Then think about the question. If the question asks for the greatest number of items you can buy, round down. If it asks for the fewest hours needed, round up. If x must be a whole number, round down for “at most” and round up for “at least”. Always check that your final whole number works in the original inequality.
Inequalities in two variables
The graph of y ≤ −x + 4 is a half-plane. First draw the boundary line y = −x + 4. Use a solid line for ≤ or ≥, and a dashed line for < or > (points on a dashed line are not solutions). Then pick a test point that is not on the line, such as (0, 0). Put it into the inequality: 0 ≤ 4 is true, so shade the side that contains (0, 0). With y ≤ the shading is below the line, and with y ≥ it is above the line, when y is alone on the left.
Systems of inequalities
A point is a solution of a system only if it makes every inequality true. The solution region is where the shaded half-planes overlap. On the SAT you usually get four points: put each into each inequality and cross out a point as soon as it fails one. Pay attention to points on a dashed boundary, such as y > 2x − 3 with the point (4, 5): 5 > 5 is false.
| Point | x + y ≤ 6 | y > x − 2 | In both? |
|---|---|---|---|
| (1, 2) | 3 ≤ 6 true | 2 > −1 true | yes |
| (5, 4) | 9 ≤ 6 false | 4 > 3 true | no |
| (0, −3) | −3 ≤ 6 true | −3 > −2 false | no |
- Distribute: −2x + 6 ≥ 4x + 18.
- Subtract 4x from both sides: −6x + 6 ≥ 18.
- Subtract 6: −6x ≥ 12.
- Divide by −6 and flip the sign: x ≤ −2.
- Check with x = −3: −2(−6) = 12 and 4(−3) + 18 = 6, and 12 ≥ 6 ✓. Check with x = 0: 6 ≥ 18 is false ✓.
- Let n be the number of banners. The cost is 12n + 18, and it must be at most 160.
- 12n + 18 ≤ 160
- 12n ≤ 142, so n ≤ 11.83…
- The club cannot buy part of a banner, and it cannot go over the budget, so round down: n = 11.
- Check: 12(11) + 18 = 150 ≤ 160 ✓. For 12 banners: 162 > 160 ✗.
- Test (0, 3): 3 ≥ 1 is true, and 3 < 5 is true. Both hold ✓.
- Test (3, 5): 5 ≥ 4 is true, but 5 < 2 is false. ✗
- Test (4, 2): 2 ≥ 5 is false. ✗
- Test (−1, −2): −2 ≥ 0 is false. ✗
- Only (0, 3) is a solution. It lies inside the shaded triangle.
- Subtract 5: −3x < −15.
- Divide by −3 and flip the sign: x > 5.
- x must be greater than 5, but not equal to 5. The least integer is 6.
- Check: 5 − 3(6) = −13 < −10 ✓. For x = 5: 5 − 15 = −10, which is not less than −10 ✗.
- Forgetting to flip the signDividing or multiplying by a negative number turns < into > and ≤ into ≥. Test one number from your answer in the original inequality.
- Flipping when you should notAdding, subtracting, or dividing by a positive number never changes the sign. Only a negative multiplier or divisor does.
- Mixing up “at least” and “at most”At least = ≥ (a floor). At most = ≤ (a ceiling). Say the story out loud: “the cost cannot go above 160” means cost ≤ 160.
- Rounding the wrong wayGreatest number you can afford: round down. Fewest needed to reach a goal: round up. Put your whole number back into the inequality.
- Counting a boundary point on a dashed lineWith < or > the boundary is not included. A point that gives 5 > 5 is not a solution.
- Checking a point in only one inequalityFor a system a point must make every inequality true. Cross out a point at the first failure.
Desmos draws inequalities. Type y ≤ −x + 4 (use the < and > keys, or type <= and >=) and the correct half-plane is shaded, with a solid or dashed boundary. Type a second inequality to see the overlap, which is the solution of a system. Then type each answer-choice point, for example (1, 0), and see if it lies inside the shaded area. One-variable inequalities also work: type −3x + 5 > 17 and Desmos shades the values of x that work.
- Type y >= x + 1
- Type y < −x + 5
- The overlap is the solution region; click the corner to read the crossing point (2, 3)
- Type (0, 3) and the other points: the one inside the overlap is the answer (points on a dashed boundary are not solutions)
Use Desmos for systems of inequalities and for checking points. For one-variable inequalities, solve by hand: it is quick, and Desmos does not warn you about flipping signs.
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