Solved Exercises on Solving Inequalities in Grade 7

Master solving inequalities: one-step, two-step, multi-step, compound inequalities, and word problems through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 One-step inequality
Exercise 1
Solve:
\(x + 7 > 12\)
Definition:

Inequality: A mathematical sentence that compares two expressions using < > ≤ ≥ ≠ symbols

One-step inequality method:
  1. Isolate the variable by performing the inverse operation
  2. Keep the inequality symbol facing the same direction
  3. Check the solution by substituting a test value
Original
\(x + 7 > 12\)
Subtract 7
\(x > 5\)
Step 1: Subtract 7 from both sides

\(x + 7 - 7 > 12 - 7\)

Step 2: Simplify

\(x > 5\)

Step 3: Check the solution

Test \(x = 6\): \(6 + 7 = 13 > 12\) ✓

\(x > 5\)
Final answer:

All values greater than 5 satisfy the inequality

Applied rules:

• Addition Property: Adding/subtracting same number preserves inequality

• Sign Direction: Inequality symbol stays the same when adding/subtracting

• Verification: Always substitute back to confirm solution

2 Two-step inequality
Exercise 2
Solve:
\(3x - 5 \leq 10\)
Definition:

Two-step inequality: Requires two operations to isolate the variable

Original
\(3x - 5 \leq 10\)
Add 5
\(3x \leq 15\)
Divide by 3
\(x \leq 5\)
Step 1: Add 5 to both sides

\(3x - 5 + 5 \leq 10 + 5\)

\(3x \leq 15\)

Step 2: Divide both sides by 3

\(\frac{3x}{3} \leq \frac{15}{3}\)

\(x \leq 5\)

Step 3: Check the solution

Test \(x = 5\): \(3(5) - 5 = 15 - 5 = 10 \leq 10\) ✓

\(x \leq 5\)
Final answer:

All values less than or equal to 5 satisfy the inequality

Applied rules:

• Two-step process: Undo addition/subtraction first, then multiplication/division

• Sign direction: Division by positive number keeps inequality symbol

• Equality allowed: \(\leq\) includes the boundary value

3 Negative coefficient
Exercise 3
Solve:
\(-2x + 8 < 4\)
Definition:

Negative coefficient: When dividing/multiplying by negative, flip inequality symbol

Original
\(-2x + 8 < 4\)
Subtract 8
\(-2x < -4\)
Divide by -2
\(x > 2\)
Step 1: Subtract 8 from both sides

\(-2x + 8 - 8 < 4 - 8\)

\(-2x < -4\)

Step 2: Divide both sides by -2 (FLIP SIGN)

\(\frac{-2x}{-2} > \frac{-4}{-2}\)

\(x > 2\)

Step 3: Check the solution

Test \(x = 3\): \(-2(3) + 8 = -6 + 8 = 2 < 4\) ✓

\(x > 2\)
Final answer:

All values greater than 2 satisfy the inequality

Applied rules:

• Negative multiplication/division: Flip inequality symbol when multiplying/dividing by negative number

• Order matters: Always divide/multiply last to apply sign rule correctly

• Memory trick: Think of inequality as a crocodile eating the larger number

Inequality Rules and Properties
\(a < b \Rightarrow a + c < b + c\)
Addition Property
Multiplication Property
\(a < b, c > 0 \Rightarrow ac < bc\)
Multiplying by positive keeps sign
Multiplication Property
\(a < b, c < 0 \Rightarrow ac > bc\)
Multiplying by negative flips sign
Transitive Property
\(a < b, b < c \Rightarrow a < c\)
Chain inequalities
Key definitions:

Inequality symbols: < (less than), > (greater than), ≤ (less than or equal), ≥ (greater than or equal)

Solution set: Set of all values that make the inequality true

Boundary point: Value where inequality changes from true to false

Complete methodology:
  1. Isolate the variable: Use inverse operations to get variable alone
  2. Watch the sign: Flip when multiplying/dividing by negative numbers
  3. Check your work: Substitute a test value to verify solution
  4. Graph the solution: Represent on number line
Tip 1: Remember to flip the inequality symbol when multiplying or dividing by a negative number!
Tip 2: Use an open circle for < and >, closed circle for ≤ and ≥ on number lines.
Tip 3: Always verify your solution by substituting a value from the solution set.
Tip 4: The solution to an inequality is often a range of values, not just one number.
Common errors: Forgetting to flip inequality symbol when multiplying/dividing by negative numbers, treating inequalities like equations.
Key memory aid: "Negative times flips the sign" - when you multiply or divide by a negative number, reverse the inequality symbol.
Properties to remember:

• Addition/Subtraction: Same operation on both sides preserves inequality

• Multiplication/Division by positive: Preserves inequality direction

• Multiplication/Division by negative: Reverses inequality direction

• Transitivity: If a < b and b < c, then a < c

Solution: Exercises 4 to 5
4 Compound inequality
Exercise 4
Solve:
\(-3 < 2x + 1 \leq 7\)
Definition:

Compound inequality: Two inequalities connected by AND or written as a chain

Original
\(-3 < 2x + 1 \leq 7\)
Subtract 1
\(-4 < 2x \leq 6\)
Divide by 2
\(-2 < x \leq 3\)
Step 1: Subtract 1 from all parts

\(-3 - 1 < 2x + 1 - 1 \leq 7 - 1\)

\(-4 < 2x \leq 6\)

Step 2: Divide all parts by 2

\(\frac{-4}{2} < \frac{2x}{2} \leq \frac{6}{2}\)

\(-2 < x \leq 3\)

Step 3: Check the solution

Test \(x = 0\): \(-3 < 2(0) + 1 = 1 \leq 7\) ✓

Test \(x = 3\): \(-3 < 2(3) + 1 = 7 \leq 7\) ✓

\(-2 < x \leq 3\)
Final answer:

All values between -2 and 3 (including 3, excluding -2) satisfy the inequality

Applied rules:

• Chain manipulation: Perform same operation on all parts of compound inequality

• AND connection: Both conditions must be satisfied simultaneously

• Bounded solution: Solution is a finite interval

5 Word problem
Exercise 5
Sarah has $50. She wants to buy books that cost $8 each. How many books can she buy if she needs to keep at least $10?
Definition:

Word problem: Translate verbal situation into mathematical inequality

Let x = books
\(50 - 8x \geq 10\)
Subtract 50
\(-8x \geq -40\)
Divide by -8
\(x \leq 5\)
Step 1: Define variable and set up inequality

Money left ≥ $10

Initial money - (cost per book × number of books) ≥ $10

\(50 - 8x \geq 10\)

Step 2: Solve the inequality

\(50 - 8x \geq 10\)

\(-8x \geq 10 - 50\)

\(-8x \geq -40\)

Step 3: Divide by -8 (FLIP SIGN)

\(\frac{-8x}{-8} \leq \frac{-40}{-8}\)

\(x \leq 5\)

Step 4: Interpret the solution

Sarah can buy at most 5 books to keep at least $10

Step 5: Verify the solution

If she buys 5 books: $50 - $8(5) = $50 - $40 = $10 ≥ $10 ✓

If she buys 6 books: $50 - $8(6) = $50 - $48 = $2 < $10 ✗

\(x \leq 5\), so Sarah can buy at most 5 books
Final answer:

Sarah can buy at most 5 books while keeping at least $10

Applied rules:

• Problem translation: Convert words to mathematical expressions

• Context consideration: Only positive integer solutions make sense

• Real-world interpretation: Check if solution makes practical sense

Inequality Laws, Methods, and Key Concepts
\(a < b \Leftrightarrow a + c < b + c\)
Addition Property
Key definitions:

Inequality: Mathematical statement comparing two expressions using <, >, ≤, ≥, or ≠

Solution set: Collection of all values that make the inequality true

Equivalent inequalities: Inequalities with the same solution set

Complete methodology:
  1. Analyze the inequality: Identify the type (one-step, two-step, etc.)
  2. Isolate the variable: Use inverse operations systematically
  3. Watch the sign: Remember to flip when multiplying/dividing by negatives
  4. Verify the solution: Check by substituting values
  5. Graph the solution: Show solution set on number line
Tip 1: When solving compound inequalities, perform operations on ALL parts simultaneously.
Tip 2: On number lines, use open circles for strict inequalities (< or >) and closed circles for inclusive inequalities (≤ or ≥).
Tip 3: Always check your solution by substituting a value from the solution set back into the original inequality.
Tip 4: In word problems, define your variable clearly and check if the solution makes practical sense.
Common errors: Forgetting to flip inequality symbol when multiplying/dividing by negative numbers, mixing up boundary points, incorrect graphing on number lines.
Exam preparation: Practice solving different types of inequalities, memorize the sign-flipping rule, work on word problems, practice number line representation.
Properties to know by heart:

• Addition Property: \(a < b \Rightarrow a + c < b + c\)

• Subtraction Property: \(a < b \Rightarrow a - c < b - c\)

• Multiplication Property (positive): \(a < b, c > 0 \Rightarrow ac < bc\)

• Multiplication Property (negative): \(a < b, c < 0 \Rightarrow ac > bc\)

• Division Property (positive): \(a < b, c > 0 \Rightarrow \frac{a}{c} < \frac{b}{c}\)

• Division Property (negative): \(a < b, c < 0 \Rightarrow \frac{a}{c} > \frac{b}{c}\)

Visualizing Inequalities: Number Line Representation
Exercise 6: Graphing Solutions
Consider the following inequalities and their solutions:
\(x > 3\) (solution: open circle at 3, arrow to the right)
\(x \leq -2\) (solution: closed circle at -2, arrow to the left)
\(-1 < x \leq 4\) (solution: open circle at -1, closed circle at 4, line between)

Analysis: The chart shows how different inequality types correspond to specific number line representations.

  • \(x > a\): Open circle at \(a\), arrow pointing right
  • \(x \geq a\): Closed circle at \(a\), arrow pointing right
  • \(x < a\): Open circle at \(a\), arrow pointing left
  • \(x \leq a\): Closed circle at \(a\), arrow pointing left
  • \(a < x < b\): Open circles at \(a\) and \(b\), line between

Questions & Answers

Question: I don't understand why I need to flip the inequality symbol when multiplying or dividing by a negative number. Can you explain why this happens?

Answer: Great question! This happens because multiplying or dividing by a negative number reverses the order of numbers on the number line.

Example: We know that 3 < 5 is true. If we multiply both sides by -1:

  • Left side: -1 × 3 = -3
  • Right side: -1 × 5 = -5
  • Now we have: -3 < -5?
  • But -3 is actually greater than -5, so -3 > -5

So 3 < 5 becomes -3 > -5 when multiplied by -1. The inequality symbol had to flip to maintain the truth of the statement.

Think of it like this: if you owe $3 (which is -3) and I owe $5 (which is -5), I owe more than you, but in the negative direction, so -3 > -5.

Question: When solving compound inequalities like \(-2 < x + 3 \leq 5\), why do I perform the same operation on all parts?

Answer: A compound inequality like \(-2 < x + 3 \leq 5\) is actually two inequalities connected by "AND":

  • \(-2 < x + 3\) (x + 3 is greater than -2)
  • \(x + 3 \leq 5\) (x + 3 is less than or equal to 5)

Both conditions must be true simultaneously. So when you subtract 3 from the middle part (x + 3), you must also subtract 3 from both the left part (-2) and the right part (5) to maintain the balance and truth of both inequalities.

If you only changed one part, you'd be altering the relationship between the parts and potentially getting an incorrect solution.

Question: How do I know whether to use an open or closed circle when graphing inequalities on a number line?

Answer: The type of circle depends on whether the boundary point is included in the solution:

  • Open circle (○): Used for strict inequalities: < and >. The boundary point is NOT included in the solution.
  • Closed circle (●): Used for inclusive inequalities: ≤ and ≥. The boundary point IS included in the solution.

Memory trick: Think of the inequality symbol as a mouth that's either "open" (○) for strict inequalities or "closed" (●) for inclusive inequalities.

Examples:

  • \(x > 3\): Open circle at 3, arrow pointing right
  • \(x \geq 3\): Closed circle at 3, arrow pointing right
  • \(x < -2\): Open circle at -2, arrow pointing left
  • \(x \leq -2\): Closed circle at -2, arrow pointing left

Question: How can I check if my solution to an inequality is correct? Sometimes I'm not sure about my answer.

Answer: There are several ways to verify your inequality solution:

Method 1: Test a value from the solution set

  • Pick a number that satisfies your solution
  • Substitute it back into the original inequality
  • If the inequality is true, your solution is likely correct

Method 2: Test a value outside the solution set

  • Pick a number that does NOT satisfy your solution
  • Substitute it back into the original inequality
  • If the inequality is false, your solution is likely correct

Method 3: Test the boundary point

  • For inequalities with ≤ or ≥, test the boundary value
  • For inequalities with < or >, the boundary value should make the original inequality false

Example: If you solved \(x + 4 > 7\) and got \(x > 3\), test \(x = 4\): \(4 + 4 = 8 > 7\) ✓

Question: What is the difference between solving inequalities and solving equations? They seem very similar.

Answer: While solving inequalities and equations share similarities, there are key differences:

Similarities:

  • Both involve isolating the variable using inverse operations
  • Both preserve equality/inequality when adding/subtracting the same value to both sides
  • Both require checking the solution

Differences:

  • Sign flipping: When multiplying or dividing by a negative number, equations stay the same but inequalities flip their symbol
  • Solutions: Equations usually have one solution, inequalities typically have a range of solutions
  • Representation: Equations show exact values, inequalities show intervals
  • Verification: Equations verify with exact substitution, inequalities verify with testing values from the solution set

Key rule to remember: The only operation that behaves differently is multiplying or dividing by a negative number - this is where inequalities require extra care compared to equations.