Algebra I · Algebra Ignite · Inequalities · Graphing
Linear Inequalities: Solving and Graphing (Theory and 30 Practice Questions)
A step-by-step guide to solving and graphing linear inequalities — the flip-the-sign rule, interval and set-builder notation, number-line graphs, compound and absolute-value inequalities, and two-variable inequalities in the coordinate plane — followed by 30 practice questions (10 easy, 10 medium, 5 medium-hard, 5 hard) with hidden step-by-step solutions. Written for Algebra I students in the SOMATH Algebra Ignite track on the Upper West Side.
Who this is for. Algebra I students who can already solve a linear equation like 3x + 5 = 20 but freeze up the first time the equals sign turns into a <, >, ≤, or ≥. Inequalities show up on almost every unit test, every Regents, every SAT, and every SHSAT. The mechanics are 95% the same as equations. The 5% that is different (the flip-the-sign rule, the two kinds of endpoints, and the shading in the plane) trips up almost every student the first time. This lesson closes that gap.
Sections
- Theory · The 4 Rules of Inequalities
- Notation · Number Lines, Intervals, Set-Builder
- Compound Inequalities (AND / OR)
- Absolute-Value Inequalities
- Two-Variable Inequalities in the Plane
- Section 1 · Easy (questions 1–10)
- Section 2 · Medium (questions 11–20)
- Section 3 · Medium-Hard (questions 21–25)
- Section 4 · Hard (questions 26–30)
- Answer Key
Theory · The 4 Rules of Inequalities
An inequality is a statement that two expressions are not equal — one is bigger, smaller, or at least as big as the other. There are four symbols you must know cold:
| Symbol | Read as | Endpoint on number line |
|---|---|---|
| < | “is less than” (strict) | Open circle ° |
| > | “is greater than” (strict) | Open circle ° |
| ≤ | “is less than or equal to” | Closed dot • |
| ≥ | “is greater than or equal to” | Closed dot • |
Rule 1 · Solve exactly like an equation — except when you multiply or divide by a negative
You can add or subtract the same number from both sides, and the inequality stays true. You can multiply or divide both sides by any positive number, and it stays true. All the moves you would use on an equation are still legal.
The one rule you cannot break. If you multiply or divide both sides of an inequality by a negative number, you must flip the inequality symbol. So < becomes >, and ≥ becomes ≤. Skipping this step is the #1 mistake on Regents inequality problems.
Why? Because 2 < 4 is true, but multiply both sides by −1 and you get −2 < −4, which is false. The flip fixes it: −2 > −4.
Rule 2 · The solution is (almost always) a set, not a single number
An equation like 2x + 3 = 11 has one answer: x = 4. An inequality like 2x + 3 > 11 has infinitely many answers: every number bigger than 4 works. Your job is to describe that entire set — not just one number in it.
The three standard ways to describe the set are: a number-line graph, interval notation, and set-builder notation. All three say the same thing. On a test, do whichever the question asks for — and if it says “solve,” do all three.
Rule 3 · Endpoints: open for strict, closed for “or equal to”
When you graph an inequality on a number line, the endpoint tells the reader whether the number itself is included. A strict inequality (< or >) gets an open circle: the endpoint is not a solution. A non-strict inequality (≤ or ≥) gets a closed dot: the endpoint is a solution. Then shade the arrow toward every number that works.
Rule 4 · Always check your answer with a test point
Pick any number in your shaded region and plug it back into the original inequality. If it makes a true statement, your answer is right. If it makes a false statement, you probably forgot to flip the sign in Rule 1.
Example. Solve −3x + 5 < 14. Subtract 5: −3x < 9. Divide by −3 and flip: x > −3. Test point: try x = 0. Original: −3(0) + 5 = 5, and 5 < 14 is true. Answer checks out.
Notation · Number Lines, Intervals, Set-Builder
Every inequality can be described three ways. Learn to translate between them fluently:
| Inequality | Number line | Interval notation | Set-builder |
|---|---|---|---|
| x < 3 | Open at 3, arrow left | (−∞, 3) | {x | x < 3} |
| x ≤ 3 | Closed at 3, arrow left | (−∞, 3] | {x | x ≤ 3} |
| x > 3 | Open at 3, arrow right | (3, ∞) | {x | x > 3} |
| x ≥ 3 | Closed at 3, arrow right | [3, ∞) | {x | x ≥ 3} |
| −2 ≤ x < 5 | Closed −2, open 5, shade between | [−2, 5) | {x | −2 ≤ x < 5} |
Bracket vs. parenthesis. A square bracket [ ] means included (matches closed dot • and matches ≤ / ≥). A round parenthesis ( ) means excluded (matches open circle ° and matches strict < / >). Infinity always gets a parenthesis, never a bracket — infinity is a direction, not a number, so you can never “include” it.
Compound Inequalities (AND / OR)
A compound inequality is two inequalities joined by AND or OR. AND means both must be true; OR means either is enough.
AND (intersection) — shade the overlap
A compound inequality like −3 < x ≤ 5 is shorthand for x > −3 AND x ≤ 5. On a number line, shade only the part that both individual inequalities agree on: the segment between −3 (open) and 5 (closed). Interval: (−3, 5].
Solving. Do the same operation to all three parts at once. Example: 1 ≤ 2x + 3 < 11. Subtract 3 from every part: −2 ≤ 2x < 8. Divide every part by 2: −1 ≤ x < 4. Interval: [−1, 4).
OR (union) — shade both pieces
A compound inequality like x < −2 OR x ≥ 4 has a solution set that is two separate pieces of the number line. You cannot write it as one interval — you have to write the union: (−∞, −2) ∪ [4, ∞).
Watch out. Do not try to combine two OR-inequalities into one three-part statement. Something like “x < −2 or x > 4” is not the same as 4 < x < −2 (which is impossible — no number is both bigger than 4 and less than −2). Always leave OR-inequalities as two pieces.
Absolute-Value Inequalities
An absolute-value inequality has the form |expression| < k or |expression| > k. Recall that |x| means the distance of x from 0 on the number line. So |x| < 3 reads “x is within 3 of zero” and |x| > 3 reads “x is more than 3 away from zero.”
“Less than” → AND (a sandwich)
If |expression| < k (with k > 0), then −k < expression < k. The solution is a single interval around zero.
Example. Solve |2x − 1| ≤ 7. Rewrite as −7 ≤ 2x − 1 ≤ 7. Add 1: −6 ≤ 2x ≤ 8. Divide by 2: −3 ≤ x ≤ 4. Interval: [−3, 4].
“Greater than” → OR (two arrows)
If |expression| > k (with k > 0), then expression < −k OR expression > k. The solution is two pieces, both going outward.
Example. Solve |3x + 2| > 8. Split: 3x + 2 < −8 OR 3x + 2 > 8. First: 3x < −10, so x < −10/3. Second: 3x > 6, so x > 2. Solution: (−∞, −10/3) ∪ (2, ∞).
Two special cases. |x| < 0 has no solution (absolute value is never negative). |x| > −5 has all real numbers as its solution (every absolute value is at least 0, which is already greater than −5). Always check the right-hand side before you split.
Two-Variable Inequalities in the Plane
A one-variable inequality (x > 3) graphs on a number line. A two-variable inequality (y > 2x + 1) graphs on a coordinate plane — the solution is a shaded half-plane bounded by a straight line.
The 4-step recipe for graphing y </> mx + b
- Solve for y if the inequality is not already in that form.
- Graph the boundary line y = mx + b using slope-intercept. Use a dashed line for strict (< or >) and a solid line for non-strict (≤ or ≥).
- Shade above or below. Above the line if y > or y ≥; below if y < or y ≤. (This works only after you solve for y first.)
- Test point. Pick a point not on the boundary — usually (0, 0) if it's not on the line — and plug into the original inequality. If it makes a true statement, you shaded correctly. If false, shade the other side.
Vertical and horizontal boundaries
If the inequality is x > 3 in the plane, the boundary is the vertical line x = 3 (dashed here), and you shade everything to the right. If it's y ≤ −2, the boundary is the horizontal line y = −2 (solid), and you shade everything below.
Systems of inequalities
Two inequalities together (a system) are graphed by shading each one separately and looking for where the shadings overlap. The overlap region is the solution set of the system — every point in it satisfies both inequalities at once. On tests, students often use two different shading colors (or diagonal hatching in opposite directions) so the overlap is easy to see.
Section 1 · Easy (questions 1–10)
One-step and two-step inequalities. Solve for x and write the solution in the form x < a, x ≥ a, etc.
Question 1 Easy
Solve for x: x + 4 > 9.
Answer: x > 5
Subtract 4 from both sides. No flip needed (we only added/subtracted). Interval: (5, ∞). Open circle at 5, arrow right.
Question 2 Easy
Solve for x: x − 7 ≤ −2.
Answer: x ≤ 5
Add 7 to both sides. Interval: (−∞, 5]. Closed dot at 5, arrow left.
Question 3 Easy
Solve for x: 3x < 15.
Answer: x < 5
Divide both sides by 3. Positive divisor — no flip. Interval: (−∞, 5).
Question 4 Easy
Solve for x: −2x < 10.
Answer: x > −5
Divide both sides by −2. Negative divisor — flip the < to >. Interval: (−5, ∞). This is the classic sign-flip problem.
Question 5 Easy
Solve for x: 2x + 5 ≥ 13.
Answer: x ≥ 4
Subtract 5: 2x ≥ 8. Divide by 2: x ≥ 4. Interval: [4, ∞).
Question 6 Easy
Solve for x: 4 − x > 1.
Answer: x < 3
Subtract 4: −x > −3. Multiply both sides by −1 and flip: x < 3. Interval: (−∞, 3).
Question 7 Easy
Solve for x: x/3 ≤ 4.
Answer: x ≤ 12
Multiply both sides by 3. Positive multiplier — no flip. Interval: (−∞, 12].
Question 8 Easy
Solve for x: −x/4 ≥ 2.
Answer: x ≤ −8
Multiply both sides by −4 and flip the ≥ to ≤. Interval: (−∞, −8].
Question 9 Easy
Solve for x: 5x − 3 < 2x + 6.
Answer: x < 3
Subtract 2x: 3x − 3 < 6. Add 3: 3x < 9. Divide by 3: x < 3. Interval: (−∞, 3).
Question 10 Easy
Write the inequality shown on this number line: closed dot at −2, arrow going right.
Answer: x ≥ −2
Closed dot → “or equal to.” Arrow right → greater than. Interval: [−2, ∞). Set-builder: {x | x ≥ −2}.
Section 2 · Medium (questions 11–20)
Multi-step, distribution, fractions, compound inequalities, and simple absolute-value.
Question 11 Medium
Solve for x: 3(x − 2) < 12.
Answer: x < 6
Distribute: 3x − 6 < 12. Add 6: 3x < 18. Divide by 3: x < 6. Interval: (−∞, 6).
Question 12 Medium
Solve for x: −4(2x + 1) ≥ 20.
Answer: x ≤ −3
Distribute: −8x − 4 ≥ 20. Add 4: −8x ≥ 24. Divide by −8 and flip: x ≤ −3. Interval: (−∞, −3].
Question 13 Medium
Solve for x: (x + 3)/2 > 5.
Answer: x > 7
Multiply both sides by 2 (positive, no flip): x + 3 > 10. Subtract 3: x > 7. Interval: (7, ∞).
Question 14 Medium
Solve for x: 6 − 2x ≤ 4x − 12.
Answer: x ≥ 3
Add 2x to both sides: 6 ≤ 6x − 12. Add 12: 18 ≤ 6x. Divide by 6: 3 ≤ x, i.e. x ≥ 3. Interval: [3, ∞).
Question 15 Medium
Solve the compound inequality: −1 < 2x + 3 ≤ 9.
Answer: −2 < x ≤ 3
Subtract 3 from every part: −4 < 2x ≤ 6. Divide every part by 2: −2 < x ≤ 3. Interval: (−2, 3]. Open circle at −2, closed dot at 3, shade between.
Question 16 Medium
Solve the compound inequality: x − 4 < −5 OR x − 4 ≥ 1.
Answer: x < −1 OR x ≥ 5
Solve each side separately. Left: x < −1. Right: x ≥ 5. Interval: (−∞, −1) ∪ [5, ∞). Two arrows going outward.
Question 17 Medium
Solve for x: |x| ≤ 6.
Answer: −6 ≤ x ≤ 6
“Less than” absolute value → sandwich. Rewrite as −6 ≤ x ≤ 6. Interval: [−6, 6].
Question 18 Medium
Solve for x: |x − 2| > 5.
Answer: x < −3 OR x > 7
“Greater than” absolute value → two arrows. Split: x − 2 < −5 OR x − 2 > 5. First: x < −3. Second: x > 7. Interval: (−∞, −3) ∪ (7, ∞).
Question 19 Medium
Write the interval-notation solution to x < 4 AND x ≥ −1.
Answer: [−1, 4)
AND → intersection. The overlap is −1 ≤ x < 4. Closed bracket at −1 (from ≥), open parenthesis at 4 (from strict <).
Question 20 Medium
Solve for x: 2x/3 − 1 < 5.
Answer: x < 9
Add 1: 2x/3 < 6. Multiply by 3/2 (positive, no flip): x < 9. Interval: (−∞, 9).
Section 3 · Medium-Hard (questions 21–25)
Fractions with LCD clearing, absolute-value with coefficient, word-problem set-up, and graphing in the coordinate plane.
Question 21 Medium-Hard
Solve for x: (x + 1)/3 − (x − 2)/4 ≥ 1.
Answer: x ≥ 2
Multiply both sides by the LCD 12 (positive, no flip): 4(x + 1) − 3(x − 2) ≥ 12. Distribute: 4x + 4 − 3x + 6 ≥ 12. Combine: x + 10 ≥ 12. Subtract 10: x ≥ 2. Interval: [2, ∞).
Question 22 Medium-Hard
Solve for x: |3x + 5| < 11.
Answer: −16/3 < x < 2
“Less than” absolute value → sandwich. Rewrite as −11 < 3x + 5 < 11. Subtract 5 from every part: −16 < 3x < 6. Divide every part by 3 (positive): −16/3 < x < 2. Interval: (−16/3, 2).
Question 23 Medium-Hard
A ride-share app charges a $3 base fee plus $2.50 per mile. Nora has at most $18 to spend. Write and solve an inequality for the greatest whole number of miles she can travel.
Answer: 6 miles
Let m = miles. Then 3 + 2.5m ≤ 18. Subtract 3: 2.5m ≤ 15. Divide by 2.5: m ≤ 6. The greatest whole-mile trip she can afford is 6 miles.
Question 24 Medium-Hard
Graph the solution of y < 2x − 3 in the coordinate plane. Describe the boundary and which region is shaded.
Answer: Dashed boundary line y = 2x − 3; shade the region below the line.
The boundary is y = 2x − 3 (slope 2, y-intercept −3). Strict < → dashed. Since the inequality is y <, shade below. Check with test point (0, 0): 0 < 2(0) − 3 = −3? No, 0 < −3 is false. So (0, 0) is not in the solution — it's above the line, and we shaded below. Correct.
Question 25 Medium-Hard
Solve for x: 5 − 3(2x − 1) > 2(x + 4) − 5.
Answer: x < 5/8
Distribute both sides: 5 − 6x + 3 > 2x + 8 − 5. Simplify: 8 − 6x > 2x + 3. Subtract 2x: 8 − 8x > 3. Subtract 8: −8x > −5. Divide by −8 and flip: x < 5/8. Interval: (−∞, 5/8).
Section 4 · Hard (questions 26–30)
Compound absolute-value, systems of inequalities, no-solution / all-real cases, and modeling.
Question 26 Hard
Solve for x: 2|x − 4| + 3 < 11.
Answer: 0 < x < 8
Isolate the absolute value first. Subtract 3: 2|x − 4| < 8. Divide by 2: |x − 4| < 4. Now the sandwich: −4 < x − 4 < 4. Add 4 to every part: 0 < x < 8. Interval: (0, 8).
Question 27 Hard
Determine all values of x that satisfy |2x + 3| ≥ −4.
Answer: All real numbers, (−∞, ∞)
Absolute value is always ≥ 0, and 0 ≥ −4 is automatically true. So every real number x satisfies this inequality. This is the “all real numbers” case — don't waste time splitting; recognize it and write the answer.
Related trap: If the problem had been |2x + 3| ≤ −4, the answer would be no solution (an absolute value is never less than a negative). Always look at the right side of an absolute-value inequality before you split.
Question 28 Hard
Solve the system in the coordinate plane and describe the solution region: y ≥ x + 1 AND y < −2x + 4.
Answer: The wedge-shaped region above the solid line y = x + 1 and below the dashed line y = −2x + 4. The two boundaries meet at (1, 2); the solution is bounded on the left by their intersection.
Line 1: y = x + 1, solid (because ≥), shade above. Line 2: y = −2x + 4, dashed (because strict <), shade below. The overlap is a wedge to the left of the intersection point. Solve the system to find the vertex: set x + 1 = −2x + 4 → 3x = 3 → x = 1, y = 2. Test point (0, 3): first, 3 ≥ 0 + 1 = 1 ✓. Second, 3 < −2(0) + 4 = 4 ✓. So (0, 3) is in the region — that confirms the wedge to the left.
Question 29 Hard
Sasha earns $12 per hour babysitting and $8 per hour dog-walking. She wants to earn at least $120 this week and can work no more than 15 hours total. Write a system of inequalities for the number of babysitting hours b and dog-walking hours d, and describe one combination that works.
Answer: 12b + 8d ≥ 120, b + d ≤ 15, b ≥ 0, d ≥ 0. One valid combination: b = 10, d = 0 (earns $120, works 10 hours).
Total pay: 12b + 8d ≥ 120 (“at least $120”). Total hours: b + d ≤ 15 (“no more than 15 hours”). Non-negativity: hours can't be negative. The solution region in the (b, d)-plane is bounded by these four inequalities. Any lattice point in the region is a valid combination. Another example: b = 6, d = 6 (pay $72 + $48 = $120 ✓, hours 12 ✓).
Question 30 Hard
Solve for x: |4x − 7| < 2x + 1.
Answer: 1 < x < 4
This is the trickiest form: the right side is a variable expression, not a constant. Two things must be true. First, 2x + 1 > 0 (otherwise the absolute value, which is ≥ 0, cannot be less than a non-positive number). That gives x > −1/2. Second, apply the “less than” sandwich: −(2x + 1) < 4x − 7 < 2x + 1.
Right piece: 4x − 7 < 2x + 1 → 2x < 8 → x < 4.
Left piece: −(2x + 1) < 4x − 7 → −2x − 1 < 4x − 7 → 6 < 6x → x > 1.
Combine with x > −1/2 from the domain check: the intersection is 1 < x < 4. Interval: (1, 4). Test point (0, 0) is not in the interval: check original — |4(0) − 7| = 7, and 2(0) + 1 = 1; 7 < 1 is false, as expected. Test x = 2 (should be in): |4(2) − 7| = 1, 2(2) + 1 = 5; 1 < 5 is true. Correct.
Answer Key
All 30 answers at a glance. Try each problem before looking here.
| # | Level | Answer |
|---|---|---|
| 1 | Easy | x > 5 |
| 2 | Easy | x ≤ 5 |
| 3 | Easy | x < 5 |
| 4 | Easy | x > −5 (flip!) |
| 5 | Easy | x ≥ 4 |
| 6 | Easy | x < 3 (flip!) |
| 7 | Easy | x ≤ 12 |
| 8 | Easy | x ≤ −8 (flip!) |
| 9 | Easy | x < 3 |
| 10 | Easy | x ≥ −2 |
| 11 | Medium | x < 6 |
| 12 | Medium | x ≤ −3 (flip!) |
| 13 | Medium | x > 7 |
| 14 | Medium | x ≥ 3 |
| 15 | Medium | −2 < x ≤ 3 |
| 16 | Medium | x < −1 OR x ≥ 5 |
| 17 | Medium | −6 ≤ x ≤ 6 |
| 18 | Medium | x < −3 OR x > 7 |
| 19 | Medium | [−1, 4) |
| 20 | Medium | x < 9 |
| 21 | Medium-Hard | x ≥ 2 |
| 22 | Medium-Hard | −16/3 < x < 2 |
| 23 | Medium-Hard | 6 miles |
| 24 | Medium-Hard | Dashed y = 2x − 3; shade below |
| 25 | Medium-Hard | x < 5/8 (flip!) |
| 26 | Hard | 0 < x < 8 |
| 27 | Hard | All real numbers |
| 28 | Hard | Wedge above solid y = x + 1 and below dashed y = −2x + 4 |
| 29 | Hard | 12b + 8d ≥ 120, b + d ≤ 15, b,d ≥ 0 |
| 30 | Hard | 1 < x < 4 |
Big Ideas Covered
- Flip on negative. Multiplying or dividing both sides by a negative number flips the inequality. Equations don't have this rule — inequalities do. Miss it once and every answer that follows is wrong.
- Open vs. closed endpoints. Strict (<, >) → open circle ° and parenthesis ( ). Non-strict (≤, ≥) → closed dot • and bracket [ ]. Infinity always gets a parenthesis.
- Interval vs. set-builder vs. number-line. All three describe the same solution set. On a test, do what's asked — if it just says “solve,” giving all three is the safest habit.
- Compound AND → a single interval (the overlap). Compound OR → a union of two pieces. Don't try to squeeze OR into one three-part statement.
- Absolute value “less than” → sandwich (AND). Absolute value “greater than” → two arrows (OR). Always isolate the absolute value first.
- Special cases. Absolute value less than a negative → no solution. Absolute value greater than a negative → all real numbers. Recognize these instantly.
- Two-variable inequalities. Solid line for ≤/≥, dashed for </>. Shade above for >, below for <. Always verify with a test point.
- Systems of inequalities. Graph each separately and find the overlap. The overlap region is the solution set.
- Test point every time. Even on easy problems, plug one number back in. It catches sign-flip errors before they compound.
- Modeling. “At most” → ≤. “At least” → ≥. “More than” / “greater than” → strict >. “Fewer than” / “less than” → strict <.