Inequalities
An inequality compares amounts that may not be equal. Instead of finding just one value, you will often find a whole range of possible values.
Foundation & Higher: symbols, number lines and linear inequalities. OCR explicitly includes chained linear inequalities at Foundation. Formal set notation, quadratic inequalities and regions in two variables are Higher.
You should know how inverse operations help solve an equation. Solving Linear Equations is the prerequisite lesson. You can also revisit Expanding Brackets when a question contains brackets.
What the symbols mean
| Symbol | Read it as | Example |
|---|---|---|
| < | less than | 2 < 5 |
| > | greater than | 7 > 4 |
| ≤ | less than or equal to | 3 ≤ 3 |
| ≥ | greater than or equal to | 8 ≥ 6 |
The open side of < or > faces the greater value. The extra line in ≤ or ≥ allows equality too.
The letter x stands for a number. Unless the question says “integer”, it may be a fraction or decimal as well as a whole number.
A solution is any value that makes the inequality true.
Worked example 1 — Read the condition
For x < 3, values such as 2, 2.9, 0 and −7 all work. The value 3 does not.
For x ≤ 3, all those values still work and 3 is included.
For x > −2, −1 is a solution because it is greater than −2. The value −3 is not.
Showing solutions on a number line
Use an open circle when the endpoint is excluded: < or >.
Use a filled circle when it is included: ≤ or ≥.
Greater values are to the right. Smaller values are to the left. An arrow shows that the range continues.
x < 3: the open circle excludes 3. The arrow means all smaller values, including values beyond the part drawn.
Worked example 2 — A range between two endpoints
The statement −2 ≤ x < 4 gives two conditions: x is at least −2 and x is less than 4.
Use a filled circle at −2, an open circle at 4 and a line between them.
The integer solutions are −2, −1, 0, 1, 2, 3. Integers are whole numbers, including negative whole numbers and zero.
Without the word “integer”, numbers such as −1.5 and 3.9 are solutions too.
−2 ≤ x < 4: include −2, exclude 4, and include every value between them.
Solving linear inequalities
Use inverse operations, just as when solving an equation. Adding or subtracting the same amount on both sides keeps their order.
Multiplying or dividing both sides by the same positive number also keeps the inequality sign.
Write the full inequality on each line. Keep the sign attached to the working.
Worked example 3 — Add, then divide
Solve 3x − 5 < 10.
Add 5 to both sides: 3x < 15.
Divide both sides by positive 3: x < 5.
Check x = 4: 12 − 5 = 7, which is less than 10. At the boundary x = 5 the result is 10, so 5 is excluded.
Checking a few values is a useful error check; the algebra establishes the whole solution range.
Worked example 4 — Brackets and an unknown on both sides
Solve 2(x + 3) ≥ x + 9.
Expand: 2x + 6 ≥ x + 9.
Subtract x: x + 6 ≥ 9.
Subtract 6: x ≥ 3.
At x = 3 both sides equal 12, which is allowed by ≥.
Multiplying or dividing by a negative number
Start with 2 < 5. Multiply both sides by −1: −2 is greater than −5, so we must write −2 > −5.
Multiplying by a negative number reverses the order. Dividing by a negative number does the same.
Reverse the inequality sign whenever you multiply or divide both sides by a negative number. This is not a rule about merely seeing a minus sign.
Worked example 5 — Divide by a negative coefficient
Solve 7 − 2x ≤ 15.
Subtract 7: −2x ≤ 8.
Divide by −2 and reverse ≤ to ≥: x ≥ −4.
Check x = 0: 7 ≤ 15 is true. Check x = −5: 17 ≤ 15 is false.
Alternatively, add 2x to both sides first: 7 ≤ 15 + 2x. Subtract 15: −8 ≤ 2x. Divide by 2: −4 ≤ x. This is the same answer.
Integer answers and real situations
Worked example 6 — Turn a budget into an inequality
A club pays £7 for delivery and £3 per badge. It can spend at most £31. Let n be the number of badges.
“At most” includes the limit: 3n + 7 ≤ 31.
Subtract 7: 3n ≤ 24.
Divide by 3: n ≤ 8.
Because badges are counted, n is a non-negative integer. The possible values are 0 to 8; the maximum is 8 badges.
Check: 8 badges cost £31, while 9 cost £34. If the club must buy at least one badge, its possible values start at 1 instead.
Translate the words carefully: “at least” means ≥, “at most” means ≤, “more than” means > and “less than” means <.
Combine linear inequalities — OCR Foundation and Higher
When two conditions are joined by and, a solution must satisfy both. Keep only their overlapping ranges.
A chained inequality is a short way to write both conditions. Apply the same operation to all three parts.
With or, satisfying either condition is enough. This matters when the solution is two separate ranges.
A solution set is the collection of allowed values. For example, {−1, 0, 1} lists three integer solutions; curly brackets group the values together.
Worked example 7 — Work on all three parts
Solve −5 < 2x + 1 ≤ 9.
Subtract 1 from all three parts: −6 < 2x ≤ 8.
Divide all three parts by positive 2: −3 < x ≤ 4.
The integer solution set is {−2, −1, 0, 1, 2, 3, 4}. The lower boundary is excluded; the upper one is included.
Worked example 8 — Two separate conditions
Solve 3x − 1 > 5 and 2x + 3 ≤ 11.
First condition: 3x > 6, so x > 2.
Second condition: 2x ≤ 8, so x ≤ 4.
Keep the overlap: 2 < x ≤ 4. The integer solutions are 3 and 4.
If the conditions had been x > 4 and x ≤ 2, there would be no overlap and therefore no solution.
Higher — quadratic inequalities
A quadratic inequality contains an x² term. First put zero on one side, then find where the quadratic equals zero. These boundary values split the number line into intervals.
Use factorising to find the boundaries, then test the sign in each interval. Do not assume that solving the corresponding equation finishes the inequality.
The Factorising lesson explains the bracket method used here.
Worked example 9 — A product is negative between its roots
Solve x² − x − 6 < 0.
Factorise: (x − 3)(x + 2) < 0.
The product is zero at x = −2 and x = 3. These are the roots, or boundary values.
Test one value in each interval. Since neither factor changes sign within an interval, each test gives the sign throughout that interval.
| Interval | Test value | Product | Sign |
|---|---|---|---|
| x < −2 | −3 | (−6)(−1) = 6 | Positive |
| −2 < x < 3 | 0 | (−3)(2) = −6 | Negative |
| x > 3 | 4 | (1)(6) = 6 | Positive |
We need a negative product, so −2 < x < 3. Exclude both roots because the inequality is strict.
For (x − 3)(x + 2) ≥ 0 instead, the answer is x ≤ −2 or x ≥ 3. The word “or” joins the two outside ranges.
Worked example 10 — Rearrange before testing
Solve x² ≥ 2x + 8.
Subtract 2x and 8: x² − 2x − 8 ≥ 0.
Factorise: (x − 4)(x + 2) ≥ 0. The roots are −2 and 4.
For x = −3 the product is 7; for x = 0 it is −8; for x = 5 it is 7.
The outside intervals are positive. Include the roots because equality is allowed.
x ≤ −2 or x ≥ 4.
Do not use “outside” as an unthinking rule: a negative leading coefficient reverses the signs. A repeated root also needs care: (x − 2)² ≥ 0 is true for every real x, while (x − 2)² ≤ 0 only allows x = 2.
Higher — graphical regions
An inequality with both x and y can describe a region on coordinate axes. Each point has an x-coordinate across and a y-coordinate up.
Draw the boundary by replacing the inequality sign with =. Use a solid line for ≤ or ≥ and a dashed line for < or >.
Test a point not on the boundary to decide which side satisfies the inequality. Show clearly whether your shading represents the allowed or excluded region; here it represents the allowed region.
For several inequalities joined by “and”, keep the overlap.
Worked example 11 — Three conditions on the same axes
Show the region satisfying x ≥ 1, y ≥ 0 and x + y < 5.
Draw x = 1 as a solid vertical line and keep the side to its right.
Draw y = 0 as a solid horizontal line and keep the side above it.
Draw x + y = 5 through (0, 5) and (5, 0) as a dashed line. The test point (0, 0) satisfies 0 + 0 < 5, so keep the side below that line.
The allowed region is the overlap shown in blue. The point (2, 1) satisfies all three conditions.
The dashed edge is excluded. In particular, (1, 4) and (5, 0) are excluded even though they also lie on solid boundaries.
Solid boundaries are included where all conditions hold. The entire dashed boundary x + y = 5 is excluded. The blue triangle shows the allowed region.
Higher — write a solution set
Set-builder notation describes all the allowed values. For example, {x : −2 < x < 3} means “the set of real numbers x such that −2 < x < 3”. A vertical bar may replace the colon. This includes fractions and decimals; it is not a list of integer answers.
For two separate ranges, {x : x ≤ −3 or x ≥ 3} means either condition may hold. Writing “and” here would demand both at once and give no solutions.
Higher practice — Write the solution of x² − 9 ≥ 0 in set-builder notation.
Factorise: (x − 3)(x + 3) ≥ 0. The product is non-negative outside the two roots, including them.
{x : x ≤ −3 or x ≥ 3}, with x real. This describes the entire solution, not just the two roots.
Think Like a Mathematician
Why does dividing by a negative reverse the sign, while subtracting a number does not?
Subtracting the same amount shifts both numbers the same distance along the number line. Their left-to-right order stays the same.
Multiplying by −1 reflects both numbers across zero. The number that was further right is now further left. That is why 2 < 5 becomes −2 > −5.
Common Mistakes
Replacing the inequality with an equals sign: keep the inequality throughout linear working. The boundary equation alone is not the whole answer.
Reversing the sign when merely subtracting: reverse it when multiplying or dividing both sides by a negative number, not whenever a minus sign appears.
Using the wrong endpoint circle: an extra line under the symbol means equality is included and the circle is filled.
Listing only integers when none were requested: an interval usually includes decimals and fractions too.
Changing only two parts of a chain: apply the same operation to all three parts.
Higher — giving only the roots: a quadratic inequality asks for intervals. Test the signs between and beyond the roots.
Higher — joining outside intervals with “and”: x < −2 or x > 3 allows either range; no number can satisfy both at once.
Higher — including a dashed boundary: strict inequalities exclude it, even at an intersection with a solid boundary.
Practice — solve and interpret
Questions 1–8 are Foundation & Higher. Questions 9–10 combine linear inequalities, including the OCR Foundation route. Questions 11–15 are Higher. Show your steps and distinguish a full range from its integer solutions.
1. Which of −1, 2, 3 and 3.5 satisfy x ≤ 3?
The sign allows any number below 3 and also 3 itself.
−1, 2 and 3 satisfy it. The number 3.5 is greater than 3.
2. Describe the number line for x > −1.
Put an open circle at −1 because equality is excluded.
Draw an arrow to the right, representing every number greater than −1.
3. List the integer solutions of −3 < x ≤ 2.
Exclude −3 and include 2. Count the whole numbers between them.
−2, −1, 0, 1, 2.
4. Solve 4x + 3 < 23.
Subtract 3: 4x < 20.
Divide by positive 4: x < 5. On a number line, use an open circle at 5 and an arrow left.
5. Solve 3(x − 2) ≥ x + 4.
Expand: 3x − 6 ≥ x + 4.
Subtract x: 2x − 6 ≥ 4.
Add 6: 2x ≥ 10.
Divide by positive 2: x ≥ 5.
6. Solve 5 − 3x > 14.
Subtract 5: −3x > 9.
Divide by −3 and reverse the sign: x < −3.
Check x = −4: 5 + 12 = 17, which is greater than 14.
7. Solve 2x + 1 ≤ 10. Give the greatest integer solution.
Subtract 1: 2x ≤ 9. Divide by 2: x ≤ 4.5.
The greatest whole number allowed is 4. The real-number solution still includes values between 4 and 4.5.
8. A trip charges £6 booking plus £4 per ticket. You can spend at most £30. Form an inequality and find the maximum number of tickets.
Let n be the non-negative integer number of tickets. 4n + 6 ≤ 30.
Subtract 6: 4n ≤ 24. Divide by 4: n ≤ 6.
The maximum is 6 tickets. Six cost £30; seven would cost £34.
9. Solve −7 ≤ 3x + 2 < 11.
Subtract 2 from all three parts: −9 ≤ 3x < 9.
Divide all three parts by 3: −3 ≤ x < 3.
Use a filled circle at −3, an open circle at 3 and the segment between them.
10. Solve 2x − 1 ≥ 3 and x + 2 < 7. List the integer solutions.
First: 2x ≥ 4, so x ≥ 2. Second: x < 5.
Their overlap is 2 ≤ x < 5.
The integer solution set is {2, 3, 4}.
11. Higher: solve x² − 2x − 8 < 0.
Factorise: (x − 4)(x + 2) < 0. The roots are −2 and 4.
Test −3, 0 and 5: the products are 7, −8 and 7.
Only the interval between the roots is negative. −2 < x < 4. Exclude the roots.
12. Higher: solve x² − 9 ≥ 0.
Factorise: (x − 3)(x + 3) ≥ 0. The roots are −3 and 3.
At −4 the product is 7; at 0 it is −9; at 4 it is 7.
Keep the outside intervals and include equality: x ≤ −3 or x ≥ 3.
13. Higher: solve 6 − x − x² > 0.
Multiply both sides by −1 and reverse the sign: x² + x − 6 < 0.
Factorise: (x + 3)(x − 2) < 0. Roots: −3 and 2.
Test −4, 0 and 3: products 6, −6 and 6.
The answer is −3 < x < 2. Checking x = 0 in the original gives 6 > 0.
14. Higher: for x ≥ 1, y ≥ 0 and x + y < 5, which of (2, 1), (1, 4), (0, 2), (3, 1) are allowed?
(2, 1): 2 ≥ 1, 1 ≥ 0 and 2 + 1 = 3 < 5, so allowed.
(1, 4): the sum is 5, so it lies on the excluded dashed boundary.
(0, 2): x = 0 fails x ≥ 1.
(3, 1): all conditions hold, since 3 + 1 = 4 < 5.
(2, 1) and (3, 1) are allowed.
15. Higher: describe how to show y > x + 1 on axes. Is (2, 4) in the region?
Draw the boundary y = x + 1 through (0, 1) and (1, 2), using a dashed line.
Test (0, 2): 2 > 1 is true. Shade the allowed side above the line.
For (2, 4), 4 > 2 + 1 is true, so it is in the region.
Frequently Asked Questions
Why can an inequality have many solutions?
It describes a comparison rather than demanding equality. For example, every number below 5 satisfies x < 5.
Does moving a term reverse the sign?
Avoid thinking of terms as moving. Add or subtract the same amount on both sides; that does not reverse the sign. Multiplying or dividing both sides by a negative number does.
Should I include the boundary?
Include it for ≤ or ≥, provided it satisfies every condition. Exclude it for < or >. Substitute the boundary to check equality.
Can I divide by an expression containing x?
Be careful: its sign may depend on x and it might be zero. The linear methods here divide by known non-zero numbers. For the quadratic examples, factorise and test intervals instead.
Do I need a graph for every inequality?
No. Algebra can find the range, and a number line can communicate it. Coordinate graphs are useful when the question describes a region with two variables.