Inequalities and Solution Sets β€” TEAS Math | Nurse.org
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Math Β· Algebra & word problems

Inequalities and Solution Sets

About 14 minutes with practiceNot marked readNot yet practiced

Estimated time includes reading and one quiz. Take the time you need.

Start here: key ideas

  • An inequality describes a set of values.
  • Open endpoints exclude; closed endpoints include.
  • Multiplying or dividing by a negative reverses order.
  • Context determines whether whole-number answers make sense.
What you’ll be able to do
  • Solve and graph one-variable inequalities.
  • Reverse the sign after negative multiplication or division.
  • Interpret bounds in context.

Solve like an equation while protecting the order relationship.

Optional review

If you’d like to review the basics, visit Solving Linear Equations. You can start this lesson without completing those first.

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A set rather than one answer

An inequality compares quantities with <, >, ≀, or β‰₯. Its solution set includes every value making it true. The boundary separates solutions from nonsolutions. An open circle excludes the boundary; a closed circle includes it.

Greater than: endpoint excluded

Number line with x > 3, an open parenthesis at 3, and a ray extending right.

Open full-size image β†—

Every value to the right of 3 satisfies x > 3, but 3 itself does not. This textbook uses a parenthesis for an excluded endpoint; an open circle means the same thing.

Greater than: endpoint excluded β€” OpenStax, via Mathematics LibreTexts. CC BY 4.0. Source image reproduced unchanged.

At least: endpoint included

Number line with x β‰₯ 3, a square bracket at 3, and a ray extending right.

Open full-size image β†—

For x β‰₯ 3, include 3 and every larger value. This textbook uses a square bracket; a filled circle is another common way to show that the endpoint is included.

At least: endpoint included β€” OpenStax, via Mathematics LibreTexts. CC BY 4.0. Source image reproduced unchanged.

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Solve and protect order

Use balanced inverse operations as with equations. One special rule preserves order: multiplying or dividing both sides by a negative reverses the sign. For βˆ’3x>12, division by βˆ’3 gives x<βˆ’4. A compound inequality such as 2≀x<5 requires both comparisons: close 2, open 5, and shade between.

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Interpret a bound

β€œAt most 40” means x≀40; β€œat least 6” means xβ‰₯6; β€œmore than 10” means x>10. Context can restrict answers to whole counts. If $7 tickets must fit a $30 budget, 7t≀30 means no more than 4 whole tickets.

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Worked examples

Worked example 1: 5xβˆ’7≀18. Add 7: 5x≀25. Divide by 5: x≀5. Close 5 and shade left.

Worked example 2: 8βˆ’2x>14. Subtract 8: βˆ’2x>6. Divide by βˆ’2 and reverse: x<βˆ’3. Testing βˆ’4 gives 16>14.

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A reliable way to reason

The reversal rule follows from number-line order. Since 2<5, multiplying both values by βˆ’1 gives βˆ’2>βˆ’5: reflection across zero reverses left and right. Adding or subtracting does not reflect the line, so it does not reverse the symbol. Keep this cause in mind instead of flipping whenever a negative merely appears.

Graphing is also a verification method. For x>3, test 4; it works, so shade toward 4 and larger values. Test the boundary separately: 3>3 is false, so use an open circle. For xβ‰€βˆ’2, βˆ’2 itself works, requiring a closed circle, and βˆ’3 confirms shading left. Context may convert a numerical boundary into a usable count. If 6 notebooks must cost less than $25, 6n<25 gives n<25/6 dollars, but if n is a count instead of price, only whole-number values would be allowed. Always define the variable before deciding whether rounding or discrete interpretation is appropriate.

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Guided reasoning and error checks

Worked context: A van can carry at most 1,200 pounds. Existing cargo weighs 375 pounds, and each crate weighs 55 pounds. If c is the number of additional crates, write 375+55c≀1,200. Subtract 375 to get 55c≀825; divide by positive 55 to get c≀15. Because c counts crates, the feasible solutions are whole numbers from 0 through 15. Substitution confirms the boundary: 375+55(15)=1,200, so 15 is included.

Optional extension: solving a compound bound

For a compound bound, solve all parts together. From βˆ’4<2x+2≀10, subtract 2 throughout: βˆ’6<2x≀8. Divide every part by positive 2: βˆ’3<x≀4. The graph opens at βˆ’3, closes at 4, and shades between. Testing 0 confirms the interior.

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Final self-check

When writing an answer from a graph, translate the endpoint and arrow separately. An open point at 2 with shading right means x>2; a closed point at 2 with shading left means x≀2. Neither the circle nor the arrow alone is enough. Read both features, then test one shaded value and one unshaded value against the original inequality.

Terms to remember

inequality
A comparison using <, >, ≀, or β‰₯.
solution set
All values that make an inequality true.
boundary
The endpoint separating solutions from nonsolutions.
open circle
A graph endpoint that is excluded.
closed circle
A graph endpoint that is included.
compound inequality
Two comparisons describing one set.

Your quick summary

  • Isolate the variable with balanced operations.
  • Reverse only after multiplying or dividing by a negative.
  • Graph the endpoint and test the shaded side.
  • Translate words such as β€œat most” and β€œat least” into inclusive bounds that fit the context.

Lesson quiz

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