A linear inequality describes a set of values rather than usually identifying one answer. Solving 2x + 1 < 9 tells you which values make the statement true. These linear inequalities worked examples focus on preserving that set while rearranging, showing endpoints on a number line and selecting integer solutions when requested.
The main rule to watch is precise: reverse the inequality sign when multiplying or dividing both sides by a negative number. Adding a negative number does not trigger the rule. Nor does the mere appearance of a minus sign somewhere in the question.
Why multiplication by a negative reverses order
Start with the true statement 2 < 5. Multiply both sides by −1. The new values are −2 and −5, and −2 is greater than −5. Therefore the corresponding true statement is −2 > −5.
Multiplication by a negative number reverses the positions relative to one another on the number line. This explains why the direction changes. Memorising “a minus means flip” is too vague and often causes errors when subtracting the same quantity from both sides.
- Adding or subtracting the same number preserves the inequality direction.
- Multiplying or dividing both sides by a positive number preserves it.
- Multiplying or dividing both sides by a negative number reverses it.
- Division by zero is never permitted.
If the sign of an expression is unknown, you cannot safely multiply or divide by it without considering cases. This lesson uses known numerical coefficients so that the direction is always determined.
Worked example 1: isolate a positive coefficient
Question: Solve 3x − 7 ≤ 11.
Add 7 to both sides to obtain 3x ≤ 18. Divide both sides by positive 3, giving x ≤ 6. No sign reversal is needed.
Check the endpoint: when x = 6, the original left-hand side is 18 − 7 = 11. Equality is permitted, so 6 belongs to the solution. Check a value from the proposed region: x = 5 gives 8 ≤ 11, which is true. A value beyond the boundary, x = 7, gives 14 ≤ 11, which is false.
On a number line, place a filled circle at 6 and shade to the left. A filled circle includes the endpoint; leftward shading shows values smaller than that endpoint.
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Boundary: 6
Worked example 2: the negative-division decision
Question: Solve 5 − 2x > 13.
- Subtract 5 from both sides: −2x > 8.
- Divide both sides by −2 and reverse the sign: x < −4.
The answer is x < −4, with an open circle at −4 and shading left. The boundary −4 gives 13 > 13, which is false because this is a strict inequality.
Test x = −5: 5 − 2(−5) = 15, and 15 > 13 is true. Test the tempting incorrect answer x > −4 with x = 0: it gives 5 > 13, which is false. A numerical counterexample quickly exposes the wrong direction.
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Boundary: −4
You could instead add 2x to both sides first, giving 5 > 13 + 2x, then subtract 13 to obtain −8 > 2x. Dividing by positive 2 gives −4 > x, which is the same statement as x < −4. Different valid routes must give the same solution set.
Worked example 3: brackets and fractions
Question: Solve (2x − 3)/4 ≥ (x + 5)/3.
The positive common denominator is 12. Multiplying both sides by 12 preserves the sign:
3(2x − 3) ≥ 4(x + 5).
Expand every term: 6x − 9 ≥ 4x + 20. Subtract 4x, then add 9, giving 2x ≥ 29. Divide by positive 2 to obtain x ≥ 29/2, or x ≥ 14.5.
At x = 14.5, the two original fractions both equal 6.5. At x = 15, the left side is 27/4 = 6.75 and the right side is 20/3, approximately 6.667, so the inequality is true.
Multiplying by a common denominator is not a reason to reverse the sign when that denominator is positive. Also remember that 3(2x − 3) contains the product −9, not −3.
Worked example 4: intersect two conditions
Question: Solve −3 < 2x + 1 ≤ 9, then list its integer solutions.
This is an “and” statement: both comparisons must hold at once. Subtract 1 from all three parts:
−4 < 2x ≤ 8.
Divide all three parts by positive 2:
−2 < x ≤ 4.
The real-number solution is the interval (−2, 4]. The round bracket excludes −2; the square bracket includes 4. The integer solutions are −1, 0, 1, 2, 3 and 4.
←────○━━━━━━●────→
Left boundary: −2; right boundary: 4
If solving the conditions separately, you would obtain x > −2 and x ≤ 4. Their overlap gives the same segment. Joining both outward regions would answer a different question.
Read the wording before selecting integers
“At least” includes the stated value, so it corresponds to ≥. “At most” corresponds to ≤. “More than” and “less than” are strict. These phrases describe numerical constraints; always define the variable before translating them.
For example, if an integer n satisfies 3 < n ≤ 7, the possibilities are 4, 5, 6 and 7. The real-number solution also includes values such as 4.2, but those are removed by the separate integer requirement.
Three practice questions with solutions
Practice 1: a negative coefficient
Solve 7 − 3x ≤ 19.
Solution: Subtract 7 to get −3x ≤ 12. Dividing by −3 reverses the sign, so x ≥ −4. The endpoint works because 7 − 3(−4) = 19. Use a filled circle at −4 and shade right.
Practice 2: clear the denominator
Solve (x − 1)/3 < 2.
Solution: Multiply by positive 3: x − 1 < 6. Add 1 to obtain x < 7. An open circle is required at 7 because substituting it gives 2 < 2, which is false.
Practice 3: combine conditions and count
Find the integers satisfying both 2x + 3 > −5 and 3x − 1 ≤ 8.
Solution: The first condition gives 2x > −8, hence x > −4. The second gives 3x ≤ 9, hence x ≤ 3. Their intersection is −4 < x ≤ 3. The seven integers are −3, −2, −1, 0, 1, 2 and 3.
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From number lines to coordinate regions
The examples above describe allowed values on a number line. For two variables, continue with the original coordinate-constraint model. Its section “Draw the allowed overlap, then calculate its corners” translates each inequality into a boundary and permitted half-plane, then finds their common feasible region. Use its feasibility practice to check a point against every condition. The later objective calculations extend the method; they are not needed merely to identify the allowed region.
Frequently asked inequality questions
Do I reverse the sign when subtracting?
No. Subtracting the same quantity from both sides preserves order. Reverse only when multiplying or dividing both sides by a known negative number.
Is 4 > x different from x < 4?
No. They describe exactly the same values. When you swap the left and right sides, rewrite the comparison so it still points toward the smaller side.
When should a number-line circle be open?
Use an open circle for a strict inequality, < or >. Use a filled circle when equality is allowed, ≤ or ≥.
Does “and” mean I combine every value from both regions?
No. “And” requires the overlap. “Or” allows values satisfying either condition, so it uses the union of the solution sets.
Can a linear inequality have no solutions?
Yes. Rearranging x + 2 < x − 1 gives 2 < −1, a false statement for every x. The solution set is empty.
Can every real number be a solution?
Yes. The inequality x + 2 > x − 1 simplifies to 2 > −1, which is always true. Every real x satisfies it.
Does testing one number prove the full answer?
No. It is a useful error check, but equivalent algebraic steps establish the whole solution set. Check the endpoint and a value on each side to catch likely mistakes.
Further Mathematics revision
The Mathematics notes hub and Mathematics topical-test collection provide existing routes to related resources. Both currently require login to view their content. The one-variable lesson and solved practice here are complete without those resources.
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