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3.6 Solve Applications with Linear Inequalities

3.6 Solve Applications with Linear Inequalities

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🔟Elementary Algebra
Unit & Topic Study Guides

Translating and Solving Linear Inequality Applications

Many real-world problems don't have a single answer. Instead, they have a range of acceptable values: you need to earn at least a certain score, spend no more than a budget, or stay under a weight limit. Linear inequalities let you model these constraints mathematically and find every value that works.

Translating Real-World Scenarios

The hardest part of these problems is turning English into math. Here's a reliable process:

  1. Read the problem and identify the unknown. Assign it a variable (like xx).
  2. Spot the constraint language. These key phrases tell you which inequality symbol to use:
PhraseSymbolExample
"at least"\geq"at least 18 years old" → x18x \geq 18
"at most" / "no more than"\leq"no more than $50" → x50x \leq 50
"more than" / "greater than">>"more than 100 tickets" → x>100x > 100
"less than" / "fewer than"<<"fewer than 30 students" → x<30x < 30
  1. Write the inequality. Build an expression on one side using the variable, and place the constraint value on the other side with the correct symbol.
  2. Check for compound inequalities. Some problems have two constraints at once. For example, "between 18 and 65 years old" translates to 18x6518 \leq x \leq 65.

A common mistake: "at least" means \geq, not >>. "At least 18" includes 18 itself.

Translation of real-world scenarios, OpenAlgebra.com: Free Algebra Study Guide & Video Tutorials: Linear Inequalities (one variable)

Solving Practical Inequalities

Solving a linear inequality works almost exactly like solving a linear equation, with one critical difference.

  1. Add or subtract the same value from both sides to move constants away from the variable.
  2. Multiply or divide both sides by the same positive value to isolate the variable.
  3. If you multiply or divide by a negative number, flip the inequality symbol. This is the rule students forget most often.

Example: A student needs to spend no more than $200 on books. She has already spent $75 and each remaining book costs $25. How many more books can she buy?

  • Set up: 75+25x20075 + 25x \leq 200
  • Subtract 75: 25x12525x \leq 125
  • Divide by 25: x5x \leq 5

She can buy at most 5 more books.

Example with a negative coefficient: Solve 3x+12>0-3x + 12 > 0

  • Subtract 12: 3x>12-3x > -12
  • Divide by 3-3 and flip the symbol: x<4x < 4
Translation of real-world scenarios, Applications of Linear Functions | Boundless Algebra

Graphing and Writing Solution Sets

Once you've solved the inequality, represent the solution on a number line:

  • Open circle (\circ) for strict inequalities (<< or >>): the endpoint is not included.
  • Closed circle (\bullet) for inclusive inequalities (\leq or \geq): the endpoint is included.
  • Shade in the direction of all values that satisfy the inequality.

Then express the solution in interval notation:

  • x5x \leq 5(,5](-\infty, 5]
  • x>4x > 4(4,)(4, \infty)
  • 18x6518 \leq x \leq 65[18,65][18, 65]

Square brackets mean the endpoint is included; parentheses mean it's not. Infinity always gets a parenthesis because you can never actually reach it.

Interpreting Inequality Solutions

Finding the mathematical answer is only half the job. You also need to make sure it fits the real-world situation.

  • Translate back to context. Don't just write x5x \leq 5. Say: "She can buy 5 or fewer additional books."
  • Check for reasonableness. If your variable represents a number of people, negative or fractional answers don't make sense. You may need to round or restrict the solution further.
  • Verify units. Your answer should use the same units the problem asked about (dollars, hours, tickets, etc.).
  • Test a value. Pick a number from your solution set and plug it back into the original inequality to confirm it works. For the book example, try x=3x = 3: 75+25(3)=15075 + 25(3) = 150, and 150200150 \leq 200 ✓.