Compound Inequalities

Grade 7 · algebra · 80 practice problems · read aloud

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Compound Inequalities: Joining Two Ideas

A compound inequality combines two simple inequalities into one statement, usually with the word "and" or "or." We use them to describe a range of values that make a situation true. For example, the temperature staying between 50° and 70° is a compound inequality.

Solving "AND" Inequalities

An "AND" inequality means a value must satisfy both inequalities at the same time. The solution is where the number lines overlap. The graph often looks like a segment between two points.

Example 1: Solve and graph -1 ≤ 2x + 3 < 7

  1. Split it into two parts: -1 ≤ 2x + 3 and 2x + 3 < 7.
  2. Solve the left part: Subtract 3: -4 ≤ 2x. Divide by 2: -2 ≤ x.
  3. Solve the right part: Subtract 3: 2x < 4. Divide by 2: x < 2.
  4. Combine: -2 ≤ x < 2. This means x is between -2 and 2, including -2.

Graph: A closed circle at -2, an open circle at 2, and a line connecting them.

Solving "OR" Inequalities

An "OR" inequality means a value must satisfy at least one of the inequalities. The solution combines both number lines. The graph often has two arrows pointing away from each other.

Example 2: Solve and graph 3x + 2 ≤ -4 OR 2x - 1 > 5

  1. Solve the first: 3x ≤ -6 → x ≤ -2.
  2. Solve the second: 2x > 6 → x > 3.
  3. Combine with OR: x ≤ -2 OR x > 3.

Graph: An arrow left from a closed circle at -2, and an arrow right from an open circle at 3.

🚨 Common Mistakes

  • Flipping the inequality sign incorrectly: Only flip it when you multiply or divide by a negative number.
  • Mixing up "AND" and "OR": "AND" needs overlap (like a sandwich). "OR" includes everything from both (like two separate options).
  • Graphing errors: Remember: ≤ or ≥ use a closed circle (•). < or > use an open circle (◦).

💡 Tips & Tricks

  • Number Line Check: Always sketch a quick number line to see if your answer makes sense.
  • The "Sandwich" Method: For "AND" statements like a < x < b, solve by doing the same operation to all three parts at once (like a sandwich).
  • Keyword Clues: "Between" usually means "AND." "Either" or "Out of range" often means "OR."

Practice Makes Perfect!

Start simple! Write your own "AND" and "OR" statements about real life (e.g., "I need more than $5 AND less than $10 for a movie"). Then, solve practice problems. Check your answers by plugging a number from your solution back into the original inequality. It should make a true statement!

Practice problems

6 of the 80, worked through step by step — try them before opening the answer.

1 3x + 7 > 16 and 2x - 5 ≤ 9

Hint: Solve each inequality separately, then find where both conditions are true simultaneously.

Show the answer

Answer: 3 < x ≤ 7

  1. Solve the first inequality 3x + 7 > 16** Subtract 7 from both sides: 3x + 7 - 7 > 16 - 7 3x > 9 Divide both sides by 3: x > 3 So from the first inequality, x must be greater than 3. --- **
  2. Solve the second inequality 2x - 5 ≤ 9** Add 5 to both sides: 2x - 5 + 5 ≤ 9 + 5 2x ≤ 14 Divide both sides by 2: x ≤ 7 So from the second inequality, x must be less than or equal to 7. --- **
  3. Combine the results** From
  4. x > 3 From
  5. x ≤ 7 We need x to satisfy both conditions at the same time. So x must be greater than 3 AND less than or equal to 7. That is written as: 3 < x ≤ 7 --- **Final Answer:** 3 < x ≤ 7

Let's solve the compound inequality step by step. We have two inequalities: 1) 3x + 7 > 16 2) 2x - 5 ≤ 9 --- **

2 2x + 5 > 11 and 3x - 4 ≤ 17

Hint: Solve each inequality separately, then find where both conditions are true simultaneously. For example, if you had y + 2 > 5 and 2y - 1 < 9, you would solve each one and look for the overlapping values.

Show the answer

Answer: 3 < x ≤ 7

  1. Solve the first inequality 2x + 5 > 11** Subtract 5 from both sides: 2x + 5 - 5 > 11 - 5 2x > 6 Divide both sides by 2: x > 3 So from the first inequality: x > 3. --- **
  2. Solve the second inequality 3x - 4 ≤ 17** Add 4 to both sides: 3x - 4 + 4 ≤ 17 + 4 3x ≤ 21 Divide both sides by 3: x ≤ 7 So from the second inequality: x ≤ 7. --- **
  3. Combine the two inequalities** From
  4. x > 3 From
  5. x ≤ 7 Putting them together: 3 < x ≤ 7 --- **
  6. Interpret the result** This means x is greater than 3 and less than or equal to 7. In interval notation, this would be (3, 7] but we write it as 3 < x ≤ 7. --- **Final Answer:** 3 < x ≤ 7

Let's solve the compound inequality step by step. We have two inequalities: 1) 2x + 5 > 11 2) 3x - 4 ≤ 17 --- **

3 2x + 5 > 11 and 3x - 4 ≤ 14

Hint: Solve each inequality separately, then find where both conditions are true simultaneously. For example, if you had y + 2 > 5 and y - 1 < 8, you would solve each one and look for the overlapping values.

Show the answer

Answer: 3 < x ≤ 6

  1. Solve the first inequality 2x + 5 > 11** Subtract 5 from both sides: 2x + 5 - 5 > 11 - 5 2x > 6 Divide both sides by 2: x > 3 So the first inequality says: x > 3 --- **
  2. Solve the second inequality 3x - 4 ≤ 14** Add 4 to both sides: 3x - 4 + 4 ≤ 14 + 4 3x ≤ 18 Divide both sides by 3: x ≤ 6 So the second inequality says: x ≤ 6 --- **
  3. Combine the solutions** From
  4. x > 3 From
  5. x ≤ 6 We need x to satisfy both conditions at the same time. So x must be greater than 3 AND less than or equal to 6. This is written as: 3 < x ≤ 6 --- **Final Answer:** 3 < x ≤ 6

Let's solve the compound inequality step by step. We have two inequalities: 1) 2x + 5 > 11 2) 3x - 4 ≤ 14 --- **

4 2(x + 3) > 8 and 3x - 5 ≤ 10

Hint: Solve each inequality separately, then find where both conditions are true simultaneously.

Show the answer

Answer: 1 < x ≤ 5

  1. Solve the first inequality** 2(x + 3) > 8 Divide both sides by 2: x + 3 > 4 Subtract 3 from both sides: x > 1 So the first inequality means: x > 1. --- **
  2. Solve the second inequality** 3x - 5 ≤ 10 Add 5 to both sides: 3x ≤ 15 Divide both sides by 3: x ≤ 5 So the second inequality means: x ≤ 5. --- **
  3. Combine the solutions** From the first inequality: x > 1 From the second inequality: x ≤ 5 We need x to satisfy both conditions at the same time. So x must be greater than 1 AND less than or equal to 5. That is: 1 < x ≤ 5 --- **Final Answer:** 1 < x ≤ 5

Let's solve the system of inequalities step by step. We have: 2(x + 3) > 8 and 3x - 5 ≤ 10 --- **

5 3(2x - 5) + 7 ≥ 4(x + 3) - 1

Hint: First distribute any numbers outside parentheses, then combine like terms on each side before isolating the variable.

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Answer: x ≥ 5.5

When solving inequalities with parentheses, distribute first to eliminate them. Then combine constant terms and variable terms separately on each side. Remember that when multiplying or dividing by a negative number, you must flip the inequality sign, but this doesn't apply to addition or subtraction.

6 2(x - 3) + 5 ≥ 9 and 3x - 7 < 11

Hint: Solve each inequality separately, then find where both conditions are true simultaneously. For example, if you had y + 2 > 4 and y - 1 < 8, you would solve each one and then look for the overlapping values.

Show the answer

Answer: 5 ≤ x < 6

  1. Solve the first inequality** 2(x - 3) + 5 ≥ 9 First, distribute the 2: 2x - 6 + 5 ≥ 9 Combine like terms: 2x - 1 ≥ 9 Add 1 to both sides: 2x ≥ 10 Divide both sides by 2: x ≥ 5 So from the first inequality: x ≥ 5. --- **
  2. Solve the second inequality** 3x - 7 < 11 Add 7 to both sides: 3x < 18 Divide both sides by 3: x < 6 So from the second inequality: x < 6. --- **
  3. Combine the solutions** From the first inequality: x ≥ 5 From the second inequality: x < 6 We need x to satisfy both conditions at the same time. So: 5 ≤ x < 6 --- **Final Answer:** 5 ≤ x < 6

Let's solve the system of inequalities step by step. We have: 1) 2(x - 3) + 5 ≥ 9 2) 3x - 7 < 11 --- **

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