Linear inequalities
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Learning objectives
What you should be able to do
Solve and represent inequalities in one or two variables; translate minimum and maximum requirements; distinguish strict from inclusive boundaries; combine conditions with “and” or “or”; enforce whole-number restrictions; and reason about feasible regions and unknown coefficient signs.
Inequalities describe allowed values, not just an equality
The symbols < and > are strict: the boundary is excluded. The symbols ≤ and ≥ include equality. A solution to an inequality makes its comparison true. The solution is often an interval or a region, not one number.
Wording | Symbol | Boundary included? |
|---|---|---|
At most; no more than | ≤ | Yes. |
At least; no less than | ≥ | Yes. |
Fewer than; less than | < | No. |
More than; greater than | > | No. |
Exactly; equal to | = | Only the equality, not a range. |
Most equation operations still work. Adding or subtracting the same quantity on both sides preserves the inequality’s direction. Multiplying or dividing by a positive number also preserves it. Multiplying or dividing by a negative number reverses it:
2 < 5 −2 > −5.
The reversal follows from reflecting the number line across zero: the point that was farther right becomes farther left.
The sign-reversal rule is about the operation
Reverse the comparison when multiplying or dividing both sides by a negative number. Do not reverse merely because the equation contains a negative term, because you subtract a positive number, or because the final answer is negative. If a divisor is an unknown parameter, its sign must be determined or treated in separate cases.
Number-line notation. A closed dot includes an endpoint; an open dot excludes it. Shading to the right means larger values; shading to the left means smaller values. Interval notation uses a bracket for inclusion and a parenthesis for exclusion. Infinity is never an included endpoint.
Endpoint style is part of the solution. A single boundary value can distinguish two otherwise similar answer choices.
Compound conditions and whole-number restrictions
An and condition means both inequalities must be true; use the intersection of their solution sets. An or condition means at least one must be true; use their union. A chained inequality such as −2 < 3x + 1 ≤ 10 is an “and” statement written compactly.
Solve a chain by treating all three parts consistently. Subtract 1 from every part, then divide every part by 3. When dividing a chain by a negative number, reverse both comparison symbols; you can then rewrite the endpoints in increasing order. Alternatively, split the chain into two inequalities and intersect their answers.
“And” does not always give a bounded interval. The requirements x > 2 and x > 5 combine to x > 5. The requirements x > 5 and x < 2 have no shared solution. Similarly, an “or” union may cover the entire number line when the two pieces overlap enough.
An equation finds a boundary, not the full solution. Solving 4x − 7
Convert a real-number bound to a permitted count
For a maximum count, choose the greatest allowed integer satisfying the upper bound. For a minimum count, choose the least allowed integer satisfying the lower bound. Do not round to the nearest integer. For n ≤ 8.4, the greatest integer is 8. For n ≥ 8.4, the least integer is 9. For n < 8, the greatest integer is 7, not 8. Check the neighboring integer directly in the original context.
The domain must justify the rounding. Counts of people, boxes, or completed items are integers. A length or time may be fractional. For real x < 3, there is no greatest real solution: every candidate less than 3 can be increased slightly. For integer x < 3, there is a greatest solution, 2.
Two-variable inequalities are half-planes
Start by graphing the boundary equation, replacing the inequality symbol with =. Use a solid boundary for ≤ or ≥, a dashed boundary for < or >. Then determine which side contains solutions.
For an isolated y, y > mx + b shades above the line and y < mx + b shades below it. For a general form such as Ax + By ≤ C, either isolate y carefully or test a point not on the boundary. The origin is convenient only when it is not on the boundary. If the test point works, shade its side; otherwise shade the other side.
For x > h or x < h, the boundary is vertical and shading is right or left, respectively. For y > k or y < k, the boundary is horizontal and shading is above or below. These cases do not require a slope formula.
Systems of inequalities and feasible regions
A system joined by “and” requires every inequality to hold. Its feasible region is the common overlap. Include nonnegativity, whole-number, time, or capacity restrictions whenever the context requires them. A point satisfying a budget but violating a minimum quantity is not feasible.
Only the common region satisfies all three constraints. Every displayed boundary is solid because each comparison includes equality.
A lower bound must not exceed an upper bound. If a system requires y ≥ L(x) and y ≤ U(x), an allowed y exists exactly when L(x) ≤ U(x). If one of these bounds is strict, then at that edge you generally need L(x) < U(x). Equality leaves no room between a strict lower or upper bound and the opposite bound.
Boundary intersections are candidates, not automatic answers. An intersection may violate a third inequality or fall outside the context. Substitute it into all constraints. A strict inequality may exclude the very point that would otherwise produce a largest or smallest value.
To prove an extremum, show a bound and show it can be reached
Bound: derive a value that no feasible solution can exceed, or go below.
Attainment: exhibit a feasible point that reaches it. A plausible boundary calculation alone does not prove that the candidate is possible. If the boundary is excluded, the best value may not be attained at all.
Combine constraints when it helps. If x + y ≥ 8 and 2x + y ≥ 11, adding gives 3x + 2y ≥ 19. This directly bounds the target expression. You need not learn a general optimization algorithm to use valid additions or positive multiples of inequalities. Keep directions consistent; multiplying by a negative reverses the relation.
Parameters, endpoint checks, and interpretation
An unknown coefficient in an inequality introduces a sign question. For
ax ≤ b,
there are three cases:
Coefficient | Valid conclusion | Reason |
|---|---|---|
a > 0 | x ≤ b/a | Division by a positive value preserves direction. |
a < 0 | x ≥ b/a | Division by a negative value reverses direction. |
a | Test 0 ≤ b | If true, every real x works; if false, none works. |
Do not silently divide by an unknown sign. An expression such as k − 3 can be positive, negative, or zero. A condition like “the solution set is x ≥ 1” may be asking which sign makes the desired direction appear. The zero case must be checked separately because it can produce every real number or no solution, not the same half-line.
A parameterized point is still an ordered pair. If the candidate is (a, 2a − 1), substitute x
Three fast checks catch different errors. An endpoint check distinguishes strict from inclusive. A test value on the proposed allowed side checks direction. A context check enforces the domain, such as a whole number of boxes. These checks supplement one another; no single test does all three jobs.
A complete inequality answer has three ingredients
Relation: the correct lower bound, upper bound, interval, or region.
Endpoint rule: whether equality is permitted.
Domain: real numbers, integers, nonnegative values, or other stated restrictions. The final response may ask for one number, such as a maximum count. Your reasoning still needs all three ingredients.
Calculator check. Entering both inequalities can visualize an overlap, but shading may be clipped by the current graph window. It may also obscure a dashed boundary or an isolated endpoint. Confirm a proposed point by direct substitution, and use algebra for exact maximum or minimum claims.
Connecting back to equations. A linear equation describes equality; an inequality extends that boundary to one side. A system of equations finds common boundaries; a system of inequalities finds allowed regions. The arithmetic tools remain familiar, but signs, endpoints, and feasibility now become part of the answer.
15 worked examples
5.01. Solve an inclusive upper bound
Which inequality is equivalent to 4x − 7 ≤ 13?
A) x < 5 B) x ≤ 5 C) x ≥ 5 D) x ≤ 20
Show worked solutionHide worked solution for example 5.01
Recognize the structure. Add 7, then divide by the positive coefficient 4.
Work it through. We obtain 4x ≤ 20, so x ≤ 5. Division by a positive number preserves the direction.
Answer: B, x ≤ 5.
Check. At the endpoint, 4(5) − 7
Avoid the trap. Replacing ≤ with < incorrectly discards a valid endpoint. The boundary value is part of the solution.
5.02. Reverse the direction only when the operation requires it
Solve −3(2x − 5) > 21.
Show worked solutionHide worked solution for example 5.02
Recognize the structure. After distributing and collecting, dividing by a negative coefficient reverses the inequality.
Work it through. Expand to −6x + 15 > 21, so −6x > 6. Divide by −6:
x < −1.
Answer: x < −1.
Check. x
Avoid the trap. The reversal happens at division by −6. It is not caused merely by moving 15 to the other side.
5.03. Keep the two endpoint rules distinct
Solve the compound inequality −5 ≤ 2x + 1 < 9.
Show worked solutionHide worked solution for example 5.03
Recognize the structure. Apply the same operation to all three parts of the chain.
Work it through. Subtract 1 to get −6 ≤ 2x < 8. Divide every part by 2:
−3 ≤ x < 4.
Answer: −3 ≤ x < 4, or [−3, 4).
Avoid the trap. The left endpoint is included, but the right endpoint is excluded. Do not make both dots open or both dots closed.
5.04. An “or” statement combines allowed regions
Solve 3x + 2 < −7 or 2x − 1 ≥ 7.
Show worked solutionHide worked solution for example 5.04
Recognize the structure. Solve each inequality separately, then keep values satisfying at least one.
Work it through. The first gives 3x < −9, so x < −3. The second gives 2x ≥ 8, so x ≥ 4. Because the connector is “or,” the result is the union of these two sets.
Answer: x < −3 or x ≥ 4. In interval notation: (−∞, −3) ∪ [4, ∞).
Check. x
Avoid the trap. Using “and” would incorrectly demand that the same number be both less than −3 and at least 4. That intersection is empty.
5.05. A budget gives a maximum whole-number count
A class has $80 to spend on an activity. There is an $18 setup fee and a charge of $6.50 per participant. What is the greatest number of participants the budget permits?
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Recognize the structure. Total cost must be at most 80, and the participant count is a nonnegative integer.
Work it through. Let n be the number of participants. Then
18 + 6.50n ≤ 80 n ≤ = .
Since 124/13 is between 9 and 10, the greatest allowed integer is 9.
Answer: 9 participants.
Check. Nine participants cost $76.50; ten cost $83, exceeding the budget.
Avoid the trap. Do not round the bound to the nearest integer. The count must stay on the permitted side of the inequality.
5.06. A minimum requirement may demand rounding upward
A group needs at least 260 pencils. It already has 35 pencils and can buy boxes containing 18 pencils each. What is the least number of boxes it must buy?
Show worked solutionHide worked solution for example 5.06
Recognize the structure. The total must reach or exceed 260; the number of boxes is an integer.
Work it through. Let b be the number of boxes. The requirement is
35 + 18b ≥ 260 18b ≥ 225 b ≥ 12.5.
The least integer satisfying this is 13.
Answer: 13 boxes.
Check. Twelve boxes give 35 + 216
Avoid the trap. Buying 12.5 boxes is not permitted by the context. Rounding downward would violate the minimum requirement.
5.07. A feasible point must satisfy every condition
Which ordered pair satisfies both 2x + 3y ≤ 12 and x ≥ 1?
A) (0, 3) B) (3, 3) C) (1, 3) D) (4, 2)
Show worked solutionHide worked solution for example 5.07
Recognize the structure. Test the coordinate restriction and the total inequality for each candidate.
Work it through. For (1, 3), x = 1 ≥ 1 and 2(1) + 3(3) = 11 ≤ 12. The pair (0, 3) fails x ≥ 1. The other two give left sides 15 and 14, both too large.
Answer: C, (1, 3).
Avoid the trap. Passing one condition is not enough. In a system, the word “both” is essential.
Key takeaway. An easy variable restriction can eliminate a choice before any arithmetic. But verify every remaining condition before accepting an answer.
5.08. Translate a dashed boundary and shaded side
Which inequality is represented by the shaded half-plane in the graph?
A) x + y ≥ 4 B) x + y > 4 C) x + y < 4 D) x − y > 4
Show worked solutionHide worked solution for example 5.08
Recognize the structure. The boundary goes through (0, 4) and (4, 0); it is dashed and the region above it is shaded.
Work it through. The boundary is x + y
Answer: B, x + y > 4.
5.09. A negative coefficient changes which side is shaded
Describe the graph of 3x − 2y ≤ 8 using its boundary line and shaded side.
Show worked solutionHide worked solution for example 5.09
Recognize the structure. Isolate y; dividing by −2 reverses the comparison.
Work it through. Subtract 3x to get −2y ≤ 8 − 3x. Dividing by −2 gives
y ≥ x − 4.
The boundary is y
Answer: A solid line y
Check. The origin satisfies 0 ≤ 8 and lies above the boundary’s y-intercept of −4. It belongs in the shaded region.
Avoid the trap. The original symbol ≤ does not automatically mean “below.” That interpretation works only after isolating y.
5.10. The lower bound must fit beneath the upper bound
Real numbers x and y satisfy y ≥ 2x − 1 and y ≤ −x + 8. What is the greatest possible value of x?
Show worked solutionHide worked solution for example 5.10
Recognize the structure. For an allowed y to exist, its lower bound cannot exceed its upper bound.
Work it through. Require
2x − 1 ≤ −x + 8 3x ≤ 9 x ≤ 3.
At x
Answer: 3.
Check. The point (3, 5) satisfies both inclusive inequalities. For x > 3, the lower bound would exceed the upper bound, so no y could work.
Avoid the trap. Solving the boundary intersection gives a candidate maximum. Showing that it is included and that larger values are impossible completes the argument.
5.11. Combine a minimum quantity and a maximum cost
A club buys notebooks at $3 each and binders at $5 each. It must buy at least 20 items and spend no more than $75. What is the greatest number of binders it can buy?
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Recognize the structure. Use the item requirement to place a lower bound on cost for any chosen binder count.
Work it through. Let n count notebooks and b count binders. The constraints are n + b ≥ 20 and 3n + 5b ≤ 75. Since 3n + 5b = 3(n + b) + 2b ≥ 60 + 2b,
we need 60 + 2b ≤ 75, giving b ≤ 7.5. The integer maximum is at most 7. It is attainable with b
Answer: 7 binders.
Avoid the trap. Using only the budget would permit 15 binders, but that fails the 20-item minimum. Both conditions must constrain the answer.
5.12. Substitute a parameterized point into each inequality
The point (a, 2a − 1) satisfies x + y ≤ 11 and 2x − y > 0. What is the greatest possible value of a?
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Recognize the structure. Replace x with a and y with the entire expression 2a − 1.
Work it through. The first inequality becomes
a + (2a − 1) ≤ 11 3a ≤ 12 a ≤ 4.
The second becomes 2a − (2a − 1) = 1 > 0, true for every real a. It imposes no additional restriction. At a
Answer: 4.
Avoid the trap. Do not turn the always-true second inequality into a restriction on a. Also distribute the subtraction through both terms of 2a − 1.
5.13. An unknown sign determines the solution direction
For which values of k is the solution set of (k − 3)x ≤ k − 3 exactly x ≥ 1?
A) k > 3 B) k < 3 C) k = 3 D) All real k
Show worked solutionHide worked solution for example 5.13
Recognize the structure. The sign of k − 3 controls the effect of division, and zero is a separate case.
Work it through. If k < 3, then k − 3 < 0, so dividing reverses the relation and gives x ≥ 1. If k > 3, division preserves the direction and gives x ≤ 1. If k
Answer: B, k < 3.
Avoid the trap. Dividing by k − 3 without knowing its sign misses both the reversal and the zero case. The word “exactly” excludes the all-real solution set when k
5.14. Prove a minimum by adding constraints
Nonnegative real numbers x and y satisfy x + y ≥ 8 and 2x + y ≥ 11. What is the least possible value of 3x + 2y?
Show worked solutionHide worked solution for example 5.14
Recognize the structure. The target is exactly the sum of the two constrained expressions.
Work it through. Add the inequalities to get
3x + 2y = (x + y) + (2x + y) ≥ 8 + 11 = 19.
Now test whether equality is possible. The boundary equations x + y
Answer: 19.
Key takeaway. The lower bound and a feasible point achieving it establish the minimum. Finding one small value without a bound would not prove that no smaller value exists.
5.15. A strict boundary excludes the apparent maximum
A nonnegative integer x and a real number y satisfy y > 2x + 1 and y ≤ −x + 10. What is the greatest possible value of x?
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Recognize the structure. The strict lower bound must be strictly below the upper bound for any y to fit.
Work it through. The necessary condition is
2x + 1 < −x + 10 3x < 9 x < 3.
The greatest nonnegative integer less than 3 is 2. It works: with x
Answer: 2.
Check. At x
Avoid the trap. The intersection of the boundary lines has x
Mastery check and error prevention
Attempt the exit check below, then reveal each answer and explanation.
Before leaving this section
Check the operation that controls direction, the symbol that controls the endpoint, and the context that controls the domain. For a maximum or minimum, derive the bound and demonstrate a feasible value that reaches it.