Create Inequalities

Grade 9 · algebra · 100 practice problems · read aloud

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Creating Inequalities in Algebra

Inequalities are mathematical statements that compare two expressions using symbols like <, >, ≤, or ≥. They're essential for representing real-world situations where quantities aren't equal, like budgeting, speed limits, or minimum requirements.

How to Create Inequalities: Step-by-Step

  1. Identify the relationship: Determine what's being compared
  2. Choose the correct symbol: < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to)
  3. Write the expressions: Place the smaller quantity on the left for < or ≤
  4. Include units when applicable: Add context to your inequality

Visual Examples

Example 1: Budget Constraint

You have $50 to spend. Each shirt costs $12. How many shirts can you buy?

Solution: 12x ≤ 50
Where x = number of shirts

Example 2: Age Requirement

You must be at least 16 to get a driver's permit.

Solution: a ≥ 16
Where a = age in years

Example 3: Temperature Range

The pool is closed when temperature is below 65°F.

Solution: t > 65
Where t = temperature in °F

Common Mistakes to Avoid

🚫 Confusing inequality symbols: Remember the "alligator eats the bigger number" - the open mouth faces the larger quantity.

🚫 Misreading word problems: "At least" means ≥, "at most" means ≤, "more than" means >, "less than" means <.

🚫 Forgetting to flip the inequality: When multiplying or dividing by a negative number, you MUST reverse the inequality symbol.

Tips & Tricks

💡 Memory aid: The inequality symbol points to the smaller number, like an arrow.

💡 Number line visualization: Draw a number line to see which values satisfy your inequality.

💡 Keyword chart: Create a quick reference for common inequality words:

  • "At least" or "minimum" → ≥
  • "At most" or "maximum" → ≤
  • "More than" or "above" → >
  • "Less than" or "below" → <

Practice Suggestions

To master creating inequalities:

  • Create inequalities from real-life scenarios (allowance, phone data limits, gaming time)
  • Practice with word problems from your textbook
  • Work with a partner - one creates the scenario, the other writes the inequality
  • Use online inequality generators for instant feedback
  • Draw number lines to visualize your solutions

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 |2x - 5| < 7

Hint: Absolute value inequalities can be rewritten as compound inequalities. For |expression| < k, it means the expression is between -k and k.

Show the answer

Answer: -1 < x < 6

  1. Rewrite the absolute value inequality as a compound inequality: -7 < 2x - 5 < 7
  2. Add 5 to all three parts: -7 + 5 < 2x - 5 + 5 < 7 + 5
  3. Simplify: -2 < 2x < 12
  4. Divide all parts by 2: -2/2 < 2x/2 < 12/2
  5. Simplify: -1 < x < 6 The solution is -1 < x < 6.

2 2x² - 8x + 6 > 0

Hint: To solve quadratic inequalities, first find the roots by setting the expression equal to zero. Then determine the intervals where the quadratic expression is positive based on the parabola's direction.

Show the answer

Answer: x < 1 or x > 3

  1. Factor out the greatest common factor** All coefficients (2, -8, 6) are divisible by 2. Factor 2 out: 2 ( x² - 4x + 3 ) > 0 --- **
  2. Solve the equation inside the parentheses** We first solve the equation x² - 4x + 3 = 0 to find the boundary points. x² - 4x + 3 = 0 Factor: (x - 1)(x - 3) = 0 So x = 1 or x = 3. --- **
  3. Understand the inequality** We have 2 (x - 1)(x - 3) > 0. Since 2 > 0, the sign of the expression depends only on (x - 1)(x - 3). So we need (x - 1)(x - 3) > 0. --- **
  4. Determine where the product is positive** The zeros are x = 1 and x = 3. These divide the number line into three intervals: Interval 1: x < 1 Interval 2: 1 < x < 3 Interval 3: x > 3 Test a number in each interval: - For x < 1, pick x = 0: (0 - 1)(0 - 3) = (-1)(-3) = 3 > 0 → True - For 1 < x < 3, pick x = 2: (2 - 1)(2 - 3) = (1)(-1) = -1 < 0 → False - For x > 3, pick x = 4: (4 - 1)(4 - 3) = (3)(1) = 3 > 0 → True --- **
  5. Write the solution** The inequality is strict (> 0), so we do not include the endpoints x = 1 and x = 3. Thus the solution is: x < 1 or x > 3. --- **Final answer:** x < 1 or x > 3

Let's solve the inequality step by step. We have: 2x² - 8x + 6 > 0 --- **

3 2x² - 5x + 3 ≤ 0

Hint: To solve quadratic inequalities, first find the roots by setting the expression equal to zero, then test intervals between the roots to determine where the inequality holds true.

Show the answer

Answer: 1 ≤ x ≤ 1.5

  1. Factor the quadratic expression** We look for two numbers that multiply to 2 × 3 = 6 and add to -5. Those numbers are -2 and -3. Rewrite the middle term: 2x² - 2x - 3x + 3 ≤ 0 Group terms: (2x² - 2x) + (-3x + 3) ≤ 0 2x(x - 1) - 3(x - 1) ≤ 0 (x - 1)(2x - 3) ≤ 0 So the factored form is: (x - 1)(2x - 3) ≤ 0 --- **
  2. Find the zeros** Set each factor equal to 0: x - 1 = 0 → x = 1 2x - 3 = 0 → x = 3/2 = 1.5 These are the critical points where the expression equals 0. --- **
  3. Determine intervals to test** The critical points split the number line into three intervals: (-∞, 1), (1, 1.5), (1.5, ∞) We need to check the sign of (x - 1)(2x - 3) in each interval. - For x < 1, pick x = 0: (0 - 1)(0 - 3) = (-1)(-3) = 3 > 0 → positive - For 1 < x < 1.5, pick x = 1.2: (1.2 - 1)(2.4 - 3) = (0.2)(-0.6) = -0.12 < 0 → negative - For x > 1.5, pick x = 2: (2 - 1)(4 - 3) = (1)(1) = 1 > 0 → positive --- **
  4. Apply the inequality sign** We want (x - 1)(2x - 3) ≤ 0. That means we want where the expression is negative or zero. From the sign chart: Negative in (1, 1.5) Zero at x = 1 and x = 1.5 So the solution set is: 1 ≤ x ≤ 1.5 --- **Final answer:** 1 ≤ x ≤ 1.5

Let's solve the inequality step by step. We have: 2x² - 5x + 3 ≤ 0 --- **

4 2x² - 8x + 6 < 0

Hint: To solve quadratic inequalities, first find the roots by setting the expression equal to zero, then determine the intervals where the parabola is below the x-axis.

Show the answer

Answer: 1 < x < 3

  1. Factor out the common factor** All coefficients are divisible by 2: 2(x² - 4x + 3) < 0 --- **
  2. Factor the quadratic inside** We need two numbers that multiply to 3 and add to -4: Those numbers are -1 and -3. So: x² - 4x + 3 = (x - 1)(x - 3) Thus: 2(x - 1)(x - 3) < 0 --- **
  3. Divide both sides by 2** Since 2 > 0, dividing by 2 does not change the inequality direction: (x - 1)(x - 3) < 0 --- **
  4. Find the critical points** Set (x - 1)(x - 3) = 0 x = 1 or x = 3 These are the points where the expression equals zero. --- **
  5. Determine intervals** The critical points split the number line into three intervals: (-∞, 1), (1, 3), (3, ∞) --- **
  6. Test each interval** We want (x - 1)(x - 3) < 0 (negative). - For x in (-∞, 1): pick x = 0 (0 - 1)(0 - 3) = (-1)(-3) = 3 > 0 → positive → not part of solution. - For x in (1, 3): pick x = 2 (2 - 1)(2 - 3) = (1)(-1) = -1 < 0 → negative → satisfies inequality. - For x in (3, ∞): pick x = 4 (4 - 1)(4 - 3) = (3)(1) = 3 > 0 → positive → not part of solution. --- **
  7. Check endpoints** At x = 1 or x = 3, expression = 0, but we need < 0 (strict inequality), so endpoints are not included. --- **
  8. Write the solution** The expression is negative only on (1, 3). So: 1 < x < 3 --- **Final answer:** 1 < x < 3

Let's solve the inequality step by step. We have: 2x² - 8x + 6 < 0 --- **

5 2x² - 8x + 6 ≤ 0

Hint: To solve quadratic inequalities, first find the roots by setting the expression equal to zero, then test intervals between the roots to determine where the inequality holds true.

Show the answer

Answer: 1 ≤ x ≤ 3

  1. Factor out the common factor** All terms are divisible by 2: 2(x² - 4x + 3) ≤ 0 Divide both sides by 2 (since 2 > 0, the inequality sign stays the same): x² - 4x + 3 ≤ 0 --- **
  2. Factor the quadratic** We look for two numbers that multiply to 3 and add to -4. Those numbers are -1 and -3. So: x² - 4x + 3 = (x - 1)(x - 3) Thus: (x - 1)(x - 3) ≤ 0 --- **
  3. Find the zeros** Set (x - 1)(x - 3) = 0 x - 1 = 0 → x = 1 x - 3 = 0 → x = 3 These are the points where the expression equals zero. --- **
  4. Determine intervals** The zeros split the number line into three intervals: (-∞, 1), (1, 3), (3, ∞) We test each interval to see where (x - 1)(x - 3) is negative or positive. - For x < 1, pick x = 0: (0 - 1)(0 - 3) = (-1)(-3) = 3 > 0 → positive - For 1 < x < 3, pick x = 2: (2 - 1)(2 - 3) = (1)(-1) = -1 < 0 → negative - For x > 3, pick x = 4: (4 - 1)(4 - 3) = (3)(1) = 3 > 0 → positive --- **
  5. Apply the inequality ≤ 0** We want (x - 1)(x - 3) ≤ 0. That means we want where it is negative or zero. It is zero at x = 1 and x = 3. It is negative between 1 and 3. So the solution is: 1 ≤ x ≤ 3 --- **Final answer:** 1 ≤ x ≤ 3

Let's solve the inequality step-by-step. We have: 2x² - 8x + 6 ≤ 0 --- **

6 2x² - 8x + 6 ≥ 0

Hint: To solve quadratic inequalities, first find the roots by setting the expression equal to zero. Then determine the intervals where the parabola is above or below the x-axis based on the leading coefficient.

Show the answer

Answer: x ≤ 1 or x ≥ 3

  1. Factor out the common factor** All coefficients are divisible by 2: 2(x² - 4x + 3) ≥ 0 Divide both sides by 2 (since 2 > 0, the inequality sign stays the same): x² - 4x + 3 ≥ 0 --- **
  2. Factor the quadratic** We look for two numbers that multiply to 3 and add to -4. Those numbers are -1 and -3. So: x² - 4x + 3 = (x - 1)(x - 3) Thus the inequality becomes: (x - 1)(x - 3) ≥ 0 --- **
  3. Find the zeros** Set (x - 1)(x - 3) = 0 x - 1 = 0 → x = 1 x - 3 = 0 → x = 3 These are the points where the expression equals zero. --- **
  4. Determine intervals** The zeros split the number line into three intervals: (-∞, 1), (1, 3), (3, ∞) We test each interval to see where (x - 1)(x - 3) is ≥ 0. - For x < 1, pick x = 0: (0 - 1)(0 - 3) = (-1)(-3) = 3 > 0 → Satisfies ≥ 0 - For 1 < x < 3, pick x = 2: (2 - 1)(2 - 3) = (1)(-1) = -1 < 0 → Does not satisfy ≥ 0 - For x > 3, pick x = 4: (4 - 1)(4 - 3) = (3)(1) = 3 > 0 → Satisfies ≥ 0 --- **
  5. Include endpoints** Since the inequality is ≥ (not just >), we include x = 1 and x = 3 because they make the expression equal to 0. --- **
  6. Write the solution** The solution is: x ≤ 1 or x ≥ 3 --- **Final answer:** x ≤ 1 or x ≥ 3

Let's solve the inequality step-by-step. We have: 2x² - 8x + 6 ≥ 0 --- **

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