Solved Exercises on One-Step Inequalities in Pre-algebra

Master one-step inequalities: understanding how to solve simple inequalities with addition, subtraction, multiplication, and division through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Addition inequality
Exercise 1
Solve: \(x + 6 > 10\)
Graph the solution on a number line.
Definition:

One-step inequality: An inequality that requires only one operation to isolate the variable. The solution represents a range of values rather than a single value.

Solving method:
  1. Identify the operation performed on the variable
  2. Apply the inverse operation to both sides
  3. Keep the inequality symbol unchanged (unless multiplying/dividing by negative)
  4. Graph the solution on a number line
  5. Verify the solution with a test value
Original
\(x + 6 > 10\)
Subtract 6
\(x > 4\)
Graph
Open circle at 4, shade right
Step 1: Identify the operation

The variable \(x\) has 6 added to it, so we subtract 6 from both sides

Step 2: Apply inverse operation

Subtract 6 from both sides: \(x + 6 - 6 > 10 - 6\)

Step 3: Simplify

Left side: \(x + 6 - 6 = x\)

Right side: \(10 - 6 = 4\)

So: \(x > 4\)

Step 4: Graph the solution

Draw an open circle at 4 (since 4 is not included) and shade to the right (since \(x > 4\))

\(x > 4\)
Final answer:

\(x > 4\). The solution includes all real numbers greater than 4.

Applied rules:

• Inverse operations: Subtraction undoes addition

• Open circle: For \(>\) or \(<\) (boundary not included)

• Shading direction: Greater than means shade right

2 Subtraction inequality
Exercise 2
Solve: \(y - 3 \leq 7\)
Graph the solution.
Definition:

Inclusive inequality: An inequality using \(\leq\) or \(\geq\) symbols, where the boundary value is included in the solution set.

Original
\(y - 3 \leq 7\)
Add 3
\(y \leq 10\)
Graph
Closed circle at 10, shade left
Step 1: Identify the operation

The variable \(y\) has 3 subtracted from it, so we add 3 to both sides

Step 2: Apply inverse operation

Add 3 to both sides: \(y - 3 + 3 \leq 7 + 3\)

Step 3: Simplify

Left side: \(y - 3 + 3 = y\)

Right side: \(7 + 3 = 10\)

So: \(y \leq 10\)

Step 4: Graph the solution

Draw a closed circle at 10 (since 10 is included) and shade to the left (since \(y \leq 10\))

\(y \leq 10\)
Final answer:

\(y \leq 10\). The solution includes all real numbers less than or equal to 10.

Applied rules:

• Inverse operations: Addition undoes subtraction

• Closed circle: For \(\leq\) or \(\geq\) (boundary included)

• Shading direction: Less than means shade left

3 Multiplication inequality
Exercise 3
Solve: \(4z < 12\)
Graph the solution.
Definition:

Positive multiplication inequality: When multiplying by a positive number, the inequality direction remains unchanged.

Original
\(4z < 12\)
Divide by 4
\(z < 3\)
Graph
Open circle at 3, shade left
Step 1: Identify the operation

The variable \(z\) is multiplied by 4, so we divide both sides by 4

Step 2: Apply inverse operation

Divide both sides by 4: \(\frac{4z}{4} < \frac{12}{4}\)

Step 3: Simplify

Left side: \(\frac{4z}{4} = z\)

Right side: \(\frac{12}{4} = 3\)

So: \(z < 3\)

Step 4: Graph the solution

Draw an open circle at 3 (since 3 is not included) and shade to the left (since \(z < 3\))

\(z < 3\)
Final answer:

\(z < 3\). The solution includes all real numbers less than 3.

Applied rules:

• Inverse operations: Division undoes multiplication

• Positive multiplier: Inequality direction unchanged

• Open circle: For \(<\) (boundary not included)

Rules and methods, laws,...
\(ax + b < c\)
Linear Inequality
Addition
\(x + a < b\)
Solution: \(x < b - a\)
Subtraction
\(x - a < b\)
Solution: \(x < b + a\)
Multiplication (positive)
\(ax < b\) (where \(a > 0\))
Solution: \(x < \frac{b}{a}\)
Multiplication (negative)
\(ax < b\) (where \(a < 0\))
Solution: \(x > \frac{b}{a}\)
Key definitions:

Inequality: A mathematical statement that compares two expressions using symbols like \(<\) (less than), \(>\) (greater than), \(\leq\) (less than or equal), or \(\geq\) (greater than or equal).

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

Strict Inequality: An inequality using \(<\) or \(>\), where the boundary is not included.

Inclusive Inequality: An inequality using \(\leq\) or \(\geq\), where the boundary is included.

Sign Flip Rule: When multiplying or dividing both sides of an inequality by a negative number, the inequality symbol must be reversed.

Complete methodology:
  1. Identify the inequality symbol and type
  2. Isolate the variable using inverse operations
  3. Check for negative coefficients that might require sign flipping
  4. Simplify both sides to get the solution
  5. Graph the solution on a number line
  6. Verify the solution by testing a value
Tip 1: Always flip the inequality sign when multiplying or dividing by a negative number
Tip 2: Open circles for < and >, closed circles for ≤ and ≥
Tip 3: Less than shades left, greater than shades right
Tip 4: Test a value from your solution region to verify
Common errors: Forgetting to flip the sign when multiplying/dividing by negative numbers, using wrong circle type, shading in wrong direction, not applying operations to both sides.
Memory aids: "Flip the sign when negative", "Less than points left", "Greater than points right", "Equal means filled dot".
Solution: Exercises 4 to 5
4 Division inequality
Exercise 4
Solve: \(\frac{w}{-2} \geq 6\)
Graph the solution.
Definition:

Division by negative inequality: When dividing by a negative number, the inequality direction must be reversed.

Original
\(\frac{w}{-2} \geq 6\)
Multiply by -2 (flip sign)
\(w \leq -12\)
Graph
Closed circle at -12, shade left
Step 1: Identify the operation

The variable \(w\) is divided by \(-2\), so we multiply both sides by \(-2\)

Step 2: Apply inverse operation

Multiply both sides by \(-2\): \(\frac{w}{-2} \times (-2) \leq 6 \times (-2)\)

Step 3: Simplify and note sign flip

Left side: \(\frac{w}{-2} \times (-2) = w\)

Right side: \(6 \times (-2) = -12\)

Since we multiplied by negative, flip the inequality: \(w \leq -12\)

Step 4: Graph the solution

Draw a closed circle at \(-12\) (since \(-12\) is included) and shade to the left (since \(w \leq -12\))

\(w \leq -12\)
Final answer:

\(w \leq -12\). The solution includes all real numbers less than or equal to \(-12\).

Applied rules:

• Negative divisor: Multiplication by negative requires sign flip

• Closed circle: For \(\geq\) (boundary included)

• Shading direction: Less than means shade left

5 Multiplication by negative
Exercise 5
Solve: \(-3x > 9\)
Graph the solution.
Definition:

Negative coefficient inequality: When dividing by a negative number, the inequality direction must be reversed.

Original
\(-3x > 9\)
Divide by -3 (flip sign)
\(x < -3\)
Graph
Open circle at -3, shade left
Step 1: Identify the operation

The variable \(x\) is multiplied by \(-3\), so we divide both sides by \(-3\)

Step 2: Apply inverse operation

Divide both sides by \(-3\): \(\frac{-3x}{-3} < \frac{9}{-3}\)

Step 3: Simplify and note sign flip

Left side: \(\frac{-3x}{-3} = x\)

Right side: \(\frac{9}{-3} = -3\)

Since we divided by negative, flip the inequality: \(x < -3\)

Step 4: Graph the solution

Draw an open circle at \(-3\) (since \(-3\) is not included) and shade to the left (since \(x < -3\))

\(x < -3\)
Final answer:

\(x < -3\). The solution includes all real numbers less than \(-3\).

Applied rules:

• Negative coefficient: Division by negative requires sign flip

• Open circle: For \(>\) (boundary not included)

• Shading direction: Less than means shade left

Comprehensive Summary: One-Step Inequalities
\(ax + b < c\)
Linear Inequality
Key definitions:

One-Step Inequality: An inequality that can be solved in a single step by applying one inverse operation to both sides.

Solution Set: The collection of all values that satisfy the inequality, typically expressed as an interval.

Sign Flip Rule: When multiplying or dividing both sides of an inequality by a negative number, the inequality symbol must be reversed.

Graphing Convention: Open circles for strict inequalities (<, >), closed circles for inclusive inequalities (≤, ≥).

Complete methodology:
  1. Identify the inequality symbol and operation
  2. Apply the inverse operation to both sides
  3. Check if you're multiplying or dividing by a negative number
  4. Flip the inequality symbol if necessary
  5. Simplify to isolate the variable
  6. Graph the solution on a number line
Tip 1: Always check if the coefficient is negative before dividing/multiplying
Tip 2: Remember: positive coefficients preserve inequality direction
Tip 3: Test a value from your solution to verify correctness
Tip 4: Shading direction depends on the inequality symbol
Common errors: Forgetting to flip the inequality sign when multiplying/dividing by negative numbers, using incorrect circle type, shading in wrong direction.
Memory aids: "Flip when negative", "Less than points left", "Greater than points right".
Essential rules to remember:

• Balance principle: Perform the same operation on both sides

• Sign flip rule: Reverse inequality when multiplying/dividing by negative

• Circle convention: Open for < and >, closed for ≤ and ≥

• Shading direction: Less than shades left, greater than shades right

• Verification: Always test a solution value

Visualization: Inequality Solution Regions
Exercise 6: Multiple Inequalities
Visualizing different inequality solution regions:
\(f_1(x): x + 6 > 10\) (solution: \(x > 4\))
\(f_2(x): x - 3 \leq 7\) (solution: \(x \leq 10\))
\(f_3(x): 4x < 12\) (solution: \(x < 3\))

Analysis: The chart shows how different inequalities create different solution regions.

  • Strict inequalities create open boundaries
  • Inclusive inequalities create closed boundaries
  • Direction of shading depends on the comparison
  • Each inequality represents a half-line on the number line

Questions & Answers

Question: Why do I have to flip the inequality sign when multiplying or dividing by a negative number? I don't understand this rule.

Answer: This rule exists to preserve the truth of the inequality. Let's look at an example:

Start with: \(3 < 5\) (true)

Multiply both sides by positive 2: \(3 \times 2 < 5 \times 2\) → \(6 < 10\) (still true)

Multiply both sides by negative 2: \(3 \times (-2) < 5 \times (-2)\) → \(-6 < -10\) (false!)

We know \(-6\) is actually greater than \(-10\), so we must flip the sign:

\(3 \times (-2) > 5 \times (-2)\) → \(-6 > -10\) (true!)

Think of it this way: multiplying by a negative number reflects numbers across zero on the number line, which reverses their order. The larger number becomes smaller and vice versa.

This is the only way to maintain the logical relationship between the numbers after multiplying or dividing by a negative value.

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

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

Open circle (○): Used for strict inequalities (\(<\) or \(>\)) where the boundary point is NOT included in the solution.

  • \(x < 5\): Use open circle at 5 (5 is not part of the solution)
  • \(x > 3\): Use open circle at 3 (3 is not part of the solution)

Closed circle (●): Used for inclusive inequalities (\(\leq\) or \(\geq\)) where the boundary point IS included in the solution.

  • \(x \leq 5\): Use closed circle at 5 (5 is part of the solution)
  • \(x \geq 3\): Use closed circle at 3 (3 is part of the solution)

Memory aid: The equal sign in \(\leq\) and \(\geq\) means "includes the equal value," so we fill in the circle to show inclusion.

Question: How do I verify my solution to an inequality?

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

Method 1: Test a value from the solution region

  • Pick a value that falls within your solution range
  • Substitute it back into the original inequality
  • Check if the inequality is true

Example: For \(x + 6 > 10\), solution is \(x > 4\). Test \(x = 5\): \(5 + 6 = 11 > 10\) ✓

Method 2: Test the boundary point

  • For inclusive inequalities (\(\leq\) or \(\geq\)), the boundary should make the inequality true
  • For strict inequalities (\(<\) or \(>\)), the boundary should make the inequality false

Always verify your solution to ensure accuracy!