Math101 · Independent Ontario learning library
Math101learn.math101.caLinear Inequalities
A linear inequality in one variable compares linear expressions and has a solution set that is an interval, ray, all real numbers, or no solution.
Linear inequalities model capacity, minimum requirements, error ranges, and constraints. They also introduce set operations and boundary reasoning used throughout algebra.
Intuition and core definition
A linear inequality in one variable compares linear expressions and has a solution set that is an interval, ray, all real numbers, or no solution. Solving uses the same balance operations as linear equations, except multiplying or dividing by a negative reverses the inequality because it reverses order.
Notation, language, and conditions
Forms include $ax+b<c$, $ax+b\ge c$, and compound conditions joined by “and” or “or.” An “and” solution is an intersection satisfying both inequalities; an “or” solution is a union satisfying at least one. Interval endpoints use brackets only when finite endpoints are included; infinity always uses parentheses.
Why this idea matters
Solving a linear inequality preserves an interval of solutions, and multiplying or dividing by a negative reverses order on the number line.
A dependable method
- Distribute and combine like terms on each side.
- Move variable terms to one side and constants to the other.
- When dividing by a negative coefficient, reverse the inequality.
- For compounds, solve each part and combine by intersection or union.
- Check with boundary values and test points, then graph the complete set.
Worked example
Representations and interpretation
Algebra identifies the boundary and direction; a number-line ray displays all solutions at once. A graph of $y=2-5x$ compared with the horizontal line $y=17$ shows the same inputs where the sloping line lies above.
Reasoning about variations
If simplification removes the variable, the result determines the whole set: $2x+1<2x+5$ becomes $1<5$, true for all real $x$; $3x+4>3x+9$ becomes $4>9$, so no solution.
Common mistakes
How to check your work
- Test an interior solution and an exterior non-solution in the original inequality.
- Check endpoint inclusion by direct substitution.
- Compare the algebraic interval, number-line graph, and verbal description.
Practice
- Solve $7-2x\le15$.
- Solve $3x+2>3x-1$.
- Solve $-1<x+2\le6$.
Answers and brief solutions
Show answers
- $x\ge-4$ $-2x\le8$; dividing by $-2$ reverses the sign.
- All real numbers Subtracting $3x$ leaves the always-true statement $2>-1$.
- $-3<x\le4$ Subtract $2$ from all three parts.
Synthesis and transfer
A budget with a fixed fee and a per-use charge yields an inequality whose endpoint must be rounded down to a whole use, then checked against the original limit.
For a plan costing $18+7v$ dollars with a $60$-dollar limit, $18+7v\le60$ leads to $v\le6$. The algebraic endpoint is exactly six uses, and the surrounding integer context excludes fractional visits. If the model were rewritten as $-7v\ge-42$, dividing by $-7$ must reverse the comparison so the same interval results. Substituting one allowed value and one value just outside the boundary checks both the arithmetic and the direction of the ray. A change in the fixed fee shifts the endpoint, while a change in the per-use rate changes how quickly the budget is consumed. These parameter effects can often be predicted before solving again.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Solve $7-2x\le15$.
- $-2x\le8$; dividing by $-2$ reverses the sign.
End of lesson
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