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Linear Inequalities
Symbols, number-line graphs and step-by-step solving for grades 9–10.
A phone plan costs $20 a month plus 10 cents a text, and you can spend at most $35. How many texts can you send? The answer isn’t one number. Any number of texts from 0 up to 150 works. An inequality is how algebra describes a whole range of answers like that.
This page shows you how to read, graph and solve inequalities. Most of it works just like solving equations, with one important new rule. Every idea has a worked example, problems with answers you can reveal, and a practice machine at the end.
What is an inequality?
An equation says two things are equal. An inequality compares two things that might not be equal: one side is smaller or bigger than the other. There are four inequality symbols.
| Symbol | Read it as | Example | In words |
|---|---|---|---|
| < | is less than | x < 4 | x is less than 4 |
| > | is greater than | x > −1 | x is greater than −1 |
| ≤ | is less than or equal to | x ≤ 10 | x is at most 10 |
| ≥ | is greater than or equal to | x ≥ 16 | x is at least 16 |
To remember which is which: the open, wide side of < or > always faces the bigger number. In 3 < 8, the wide side faces 8.
The line under ≤ and ≥ is half of an equals sign. It means “or equal to,” so x ≤ 10 allows 10 itself, but x < 10 doesn’t.
Read from the variable’s side
3 < x and x > 3 say the same thing: x is bigger than 3. If you flip a whole inequality around, so the sides trade places, the symbol turns around too. It’s usually easiest to read and graph an inequality when x is on the left.
A linear inequality is like a linear equation, but with one of these four symbols instead of =. The variable only appears to the first power: 3x − 5 ≥ 7 is linear, but x² < 9 is not.
Solutions are sets
A solution of an inequality is a number that makes it true. Most linear equations have exactly one solution. Most inequalities have infinitely many.
Take x > 3. The numbers 4, 10, 3.5, 3.01 and a million are all solutions. 3 is not, because 3 is not greater than itself. The collection of every solution is called the solution set.
Here’s how to test some numbers in 2x − 1 < 5: put each one in for x and see whether the statement comes out true.
| x | Put it in | Solution? |
|---|---|---|
| −2 | 2(−2) − 1 = −5, and −5 < 5 | Yes |
| 0 | 2(0) − 1 = −1, and −1 < 5 | Yes |
| 3 | 2(3) − 1 = 5, and 5 < 5 | No: 5 is not less than 5 |
| 5 | 2(5) − 1 = 9, and 9 < 5 | No |
Solving gives the whole set at once: 2x − 1 < 5 becomes x < 3. That matches the table. −2 and 0 are below 3, while 3 and 5 are not.
Later courses sometimes write a solution set in set-builder notation: {x | x < 3} is read “the set of all x such that x is less than 3.” On this page we’ll mostly use plain inequalities, number-line graphs and interval notation.
Graphing on a number line
You can’t list infinitely many solutions, but you can draw them. A number-line graph uses two marks:
- A circle at the boundary number. It’s open (hollow) for < and >, because the boundary is not a solution. It’s closed (filled in) for ≤ and ≥, because the boundary is a solution.
- Shading over every solution. x > 3 and x ≥ 3 shade to the right, toward bigger numbers. x < 3 and x ≤ 3 shade to the left. The arrow means “and so on, forever.”
x < 3
Open circle: 3 is not included.
x ≤ 3
Closed circle: 3 is included.
x > −1
Open circle, shaded right.
x ≥ −1
Closed circle, shaded right.
Quick shading check
When x is on the left, the symbol points the way to shade: x < 3 points left, x > 3 points right. If x is on the right, as in 5 ≥ x, rewrite it first as x ≤ 5. To be sure, pick any number in the shaded part and test it.
Try it yourself. Choose a type of inequality, a symbol and a number, and the grapher draws it. Then test numbers to see which ones are solutions. You can also tap the number line to place the test point.
One-step inequalities
You solve an inequality with the same moves as an equation: add, subtract, multiply or divide both sides by the same number. The goal is still to get x alone. Adding or subtracting never changes the direction of the sign, and neither does multiplying or dividing by a positive number.
Example 1: subtract
Example 2: add
Example 3: divide
Example 4: multiply
Your turn
Solve x + 9 ≤ 4
Solve 5x > −15
Solve x/2 ≥ −3
The flip rule: multiplying or dividing by a negative
Here is the one place where inequalities behave differently from equations. Start with a true statement, 2 < 5, and do the same thing to both sides. Watch what happens to the sign.
| Do this to both sides | True statement | Sign |
|---|---|---|
| Start | 2 < 5 | < |
| Add 3 | 5 < 8 | stays < |
| Subtract 10 | −8 < −5 | stays < |
| Multiply by 2 | 4 < 10 | stays < |
| Divide by 2 | 1 < 2.5 | stays < |
| Multiply by −1 | −2 > −5 | flips to > |
| Multiply by −3 | −6 > −15 | flips to > |
| Divide by −2 | −1 > −2.5 | flips to > |
Multiplying by −1 turns 2 into −2 and 5 into −5. But −2 is greater than −5: on a number line, −2 is to the right. If you kept the old sign, you’d get −2 < −5, which is false.
Multiply or divide by a negative → flip the sign.
So < becomes >, > becomes <, ≤ becomes ≥, and ≥ becomes ≤.
Example 1
Example 2
Example 3
Check Example 1
Pick a number in x > −4, like 0: −3(0) = 0, and 0 < 12 is true. Pick one outside, like −5: −3(−5) = 15, and 15 < 12 is false. The flipped answer passes both tests.
When NOT to flip
- Adding or subtracting, even a negative number: x − 4 > −6 becomes x > −2.
- Multiplying or dividing by a positive number, even when the other side is negative: 2x < −8 becomes x < −4.
Only the sign of the number you multiply or divide by matters, not the signs of the numbers already in the inequality.
Your turn
Solve −4x ≥ 28
Solve −x/3 < 2
Solve 2 − x ≤ 7
Multi-step inequalities
Use the same order as with equations: distribute and combine like terms first, undo adding or subtracting next, then multiply or divide. Only flip at the step where you multiply or divide by a negative.
Example 1
Example 2
Example 3: distribute
Example 4: distribute a negative
Your turn
Solve 5x + 2 > −13
Solve x/4 − 1 ≤ 2
Solve 4 − 3(x + 1) ≥ 10
Variables on both sides
Collect the x-terms on one side and the numbers on the other. Here’s a handy trick: move the smaller x-term. Then the x-term that’s left has a positive coefficient, and you won’t need to flip at all.
Example 1
Example 2: x ends up on the right
You could also do Example 2 by subtracting 6x instead. You’d get −4x + 9 ≤ −3, then −4x ≤ −12, and dividing by −4 (with a flip) gives the same answer, x ≥ 3. Both ways are correct. The trick just saves you a flip.
Your turn
Solve 7x + 1 < 4x − 8
Solve x − 4 ≥ 3x + 6
Always true or never true
Sometimes the x-terms cancel and you’re left with a statement with no x at all. Then the answer depends only on whether that statement is true.
Always true
Never true
The graph of “all real numbers” is the whole line, shaded, with arrows at both ends. For “no solution” nothing is shaded.
Watch the “or equal to”
4x + 1 > 4x + 1 simplifies to 1 > 1, which is false: no solution. But 4x + 1 ≥ 4x + 1 simplifies to 1 ≥ 1, which is true: all real numbers. One small line under the sign changes the whole answer.
Compound inequalities: and, or
A compound inequality joins two inequalities with and or or.
- And means both must be true. “x is greater than −2 and at most 3” describes the numbers between −2 and 3. We write it in one chain: −2 < x ≤ 3. Read it as “x is between −2 and 3, including 3.”
- Or means at least one must be true. x < −1 or x ≥ 4 describes numbers outside the stretch from −1 to 4.
−2 < x ≤ 3
And: the part between, where both are true.
x < −1 or x ≥ 4
Or: two separate pieces.
To solve a chain like −3 < 2x + 1 ≤ 7, do the same thing to all three parts at once, so that x is alone in the middle. If you divide by a negative, flip both signs.
Example 1: and
Example 2: and, with a negative
Example 3: or
Order matters
Always write a chain from smaller to larger, like a number line: −3 < x ≤ 3, not 3 ≥ x > −3 (correct, but hard to read).
Never squash an “or” into a chain. 3 < x < −3 would mean x is bigger than 3 and smaller than −3 at the same time. No number does that.
Two surprises
An “and” can have no solution: x < 1 and x > 4 has no overlap. An “or” can cover everything: x < 5 or x > 1 includes every real number, because each number is less than 5, greater than 1, or both.
Your turn
Solve 1 ≤ x + 4 < 6
Solve −6 < 3x − 3 ≤ 9
Solve 2x − 1 ≤ −5 or x − 3 > 1
Interval notation
Interval notation is a short way to write a solution set, used a lot in later courses. You write the left end and the right end of the shaded part, separated by a comma.
- A parenthesis ( or ) means that end is not included, like an open circle.
- A bracket [ or ] means that end is included, like a closed circle.
- ∞ (infinity) and −∞ mean the shading goes on forever. They always get a parenthesis, because infinity isn’t a number you can reach.
- Write the smaller end first. Use ∪ (“union”) to join the two pieces of an “or.”
| Inequality | Interval notation | Graph |
|---|---|---|
| x > 3 | (3, ∞) | open circle at 3, shaded right |
| x ≤ −1 | (−∞, −1] | closed circle at −1, shaded left |
| −2 < x ≤ 5 | (−2, 5] | open at −2, closed at 5, shaded between |
| x < −3 or x ≥ 3 | (−∞, −3) ∪ [3, ∞) | two pieces pointing outward |
| all real numbers | (−∞, ∞) | the whole line |
| no solution | ∅ (the empty set) | nothing shaded |
Don’t mix it up with a point
(3, 5) in interval notation means every number between 3 and 5. In graphing, the same symbols name the point at x = 3, y = 5. Context tells you which one is meant.
Checking with test points
With an equation, you check one number. With an inequality, check three things:
- The boundary. Put the boundary number in. The two sides should come out equal.
- A test point inside the shaded part. It should make the original inequality true.
- A test point outside the shaded part. It should make the original inequality false.
Suppose you solved 4 − 3x ≤ 10 and got x ≥ −2. Here’s the check, always in the original inequality:
| x | Put it in 4 − 3x ≤ 10 | Result |
|---|---|---|
| −2 | 4 − 3(−2) = 10, and 10 ≤ 10 | True, and the sides are equal: −2 is the boundary, and ≤ includes it |
| 0 | 4 − 3(0) = 4, and 4 ≤ 10 | True: inside the answer, as it should be |
| −5 | 4 − 3(−5) = 19, and 19 ≤ 10 | False: outside the answer, as it should be |
Test points are great at catching a forgotten flip. If you had written x ≤ −2 instead, −5 would be inside your answer, but the table shows −5 doesn’t work. That tells you the sign points the wrong way.
Word problems: at least, at most
Word problems use phrases instead of symbols. These are the ones you’ll meet most:
| Words | Symbol | Example |
|---|---|---|
| at least, no less than, a minimum of | ≥ | You must be at least 16: a ≥ 16 |
| at most, no more than, a maximum of, up to | ≤ | At most 30 students: s ≤ 30 |
| more than, greater than, over, exceeds | > | More than 8 hours: h > 8 |
| less than, fewer than, under, below | < | Fewer than 5 mistakes: m < 5 |
Careful: “5 less than a number” is the expression n − 5, not an inequality. It’s “is less than” that means <.
The steps are the same as for equation word problems: name the variable, write the inequality, solve it, then answer the question in a sentence. Make sure the answer makes sense, since you can’t buy half a ticket or send −3 texts.
Example 1: phone plan
$20 a month plus $0.10 per text. You can spend at most $35. Let t be the number of texts.
Answer: at most 150 texts. Since t counts texts, it’s a whole number from 0 to 150.
Example 2: test average
Your scores are 78, 85 and 90. What do you need on the fourth test to average at least 85? Let s be the score.
Answer: at least 87. Tests are scored out of 100, so the real answer is 87 ≤ s ≤ 100, a compound inequality.
Example 3: elevator
An elevator can carry up to 1200 pounds. A 180-pound worker loads boxes that weigh 60 pounds each. How many boxes can go up in one trip? Let b be the number of boxes.
Answer: at most 17 boxes.
Your turn
Sam has $50 and saves $15 a week. After how many weeks will he have more than $200?
After exactly 10 weeks he has $200, which isn’t more than $200. So the answer is 11 weeks.
A taxi charges $3 plus $2.50 per mile. You want to spend no more than $18. How far can you ride?
At most 6 miles.
Common mistakes
Practice problems
Solve each one, then open the answer to check your steps and the graph. Numbers 1–4 are one step, 5–9 are multi-step, 10–11 are special cases, 12–13 are compound, and 14 is a word problem.
- x − 8 ≥ −3
Show answer
x ≥ 5Add 8 to both sides - 7x < −35
Show answer
x < −5Divide both sides by 7 (positive: no flip) - −5x ≤ 30
Show answer
x ≥ −6Divide both sides by −5 and flip ≤ to ≥ - −x/4 > 1
Show answer
x < −4Multiply both sides by −4 and flip > to < - 3x + 7 > −2
Show answer
3x > −9Subtract 7 from both sidesx > −3Divide both sides by 3 - 10 − 4x ≥ 2
Show answer
−4x ≥ −8Subtract 10 from both sidesx ≤ 2Divide both sides by −4 and flip - 2(x − 5) < 3x − 4
Show answer
2x − 10 < 3x − 4Distribute−10 < x − 4Subtract 2x (the smaller x-term)−6 < xAdd 4 to both sidesx > −6Read it from x’s side - 6x − 1 ≤ 2x + 11
Show answer
4x − 1 ≤ 11Subtract 2x from both sides4x ≤ 12Add 1 to both sidesx ≤ 3Divide both sides by 4 - x/2 + 3 > x/3 + 4
Show answer
3x + 18 > 2x + 24Multiply every term by 6 (positive: no flip)x + 18 > 24Subtract 2x from both sidesx > 6Subtract 18 from both sides - 4(x + 2) ≤ 4x + 5
Show answer
4x + 8 ≤ 4x + 5Distribute8 ≤ 5Subtract 4x: falseNo solutionNo number works - −2(x − 1) > −2x − 3
Show answer
−2x + 2 > −2x − 3Distribute −22 > −3Add 2x: trueAll real numbersEvery number works - −4 ≤ 3x + 2 < 11
Show answer
−6 ≤ 3x < 9Subtract 2 from all three parts−2 ≤ x < 3Divide all three parts by 3 - 5 − x < 2 or 2x + 4 ≤ −6
Show answer
−x < −3 or 2x ≤ −10Subtract 5 on the left, subtract 4 on the rightx > 3 or x ≤ −5Left: multiply by −1 and flip. Right: divide by 2x ≤ −5 or x > 3Write the left-most piece first - A club needs to raise at least $600. It has $150 and sells candles for $9 each. How many candles must it sell? Use c.
Show answer
150 + 9c ≥ 600“At least” means ≥9c ≥ 450Subtract 150 from both sidesc ≥ 50Divide both sides by 9At least 50 candles.
History: where do < and > come from?
The signs < and > first appeared in print in 1631, in an algebra book by the English mathematician Thomas Harriot (about 1560–1621). Its Latin title is Artis Analyticae Praxis, roughly “The Practice of the Analytic Art.” The book explains the new sign with the example a > b, meaning “a is greater than b.”
Harriot never saw it. He died in 1621, and the book was put together from his papers ten years later, mainly by his friend Walter Warner. Historians who have studied Harriot’s handwritten pages say he used different-looking signs himself. So the shapes we use today may have been chosen by the editors, not by Harriot.
Harriot did much more than algebra. He taught navigation to the sailors on Sir Walter Raleigh’s voyages, sailed to Roanoke Island in what’s now North Carolina in 1585, and on 26 July 1609 sketched the Moon through a telescope, the earliest telescope observation known in England.
And ≤ and ≥?
Those came later. The English mathematician John Wallis used a version with a single bar above the sign in 1670. The French scientist Pierre Bouguer used signs with two bars underneath, ≦ and ≧, in 1734. The single-bar ≤ and ≥ we use now became standard later still. The equals sign itself is older than all of them: Robert Recorde introduced it in 1557, as the Linear Equations history section explains.
How do we know?
The 1631 book survives and can be read today. Historians compare it with Harriot’s own manuscripts, thousands of pages of which are kept in the British Library and at Petworth House, England. They still disagree about how much of the printed book reflects Harriot’s own choices.
Try the interactive version at jugami.com/algebra/linear-inequalities/
Practice: Solve It
A new inequality every time, with hints, full steps and a graph of the answer. Choose a level, or mix them all. Type your answer like x > 3, x ≤ −2, −1 < x ≤ 4 or x < 1 or x > 5. You can type <= and >=, or tap the symbol buttons. Interval notation like (−∞, 3] works too.
Sources
- Jeff Miller, Earliest Uses of Symbols of Relation, hosted by MacTutor History of Mathematics, University of St Andrews (Harriot’s < and > in 1631, with the Latin explanation; Wallis, 1670; Bouguer, 1734; the manuscript question).
- J. J. O’Connor and E. F. Robertson, MacTutor: Thomas Harriot biography (life and dates, the 1585 voyage, the 1609 Moon sketch, the posthumous 1631 book and its editors).
- Muriel Seltman and Robert Goulding, Thomas Harriot’s Artis Analyticae Praxis: An English Translation with Commentary (Springer, 2007; publisher’s book page), whose introduction and commentary compare the 1631 book with Harriot’s surviving manuscript papers.
- Common Core State Standards for Mathematics: Reasoning with Equations and Inequalities (A-REI.B.3, solving linear inequalities in one variable) and Creating Equations (A-CED.A.1, writing inequalities for word problems), which this lesson supports.
Credits: all illustrations and number-line graphs on this page are original SVG drawings made for Jugami. Every worked example and answer was checked by testing numbers in the original inequality and in every step.