Solved Exercises on One-Step Equations in Grade 8

Master one-step equations: addition, subtraction, multiplication, division, and mixed operations through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Addition Equation
Exercise 1
Solve for \(x\):
\(x + 7 = 15\)
Definition:

One-Step Addition Equation: An equation that requires one operation to isolate the variable

Addition Method:
  1. Identify the operation affecting the variable
  2. Apply the inverse operation to both sides
  3. Simplify to find the variable
  4. Verify the solution by substituting back
Original Equation
\(x + 7 = 15\)
Subtract 7 from both sides
\(x + 7 - 7 = 15 - 7\)
Simplify
\(x = 8\)
Step 1: Identify the operation

The variable \(x\) is being increased by 7

Step 2: Apply the inverse operation

To undo adding 7, subtract 7 from both sides: \(x + 7 - 7 = 15 - 7\)

Step 3: Simplify

Left side: \(x + 7 - 7 = x\)

Right side: \(15 - 7 = 8\)

Step 4: Write the solution

\(x = 8\)

Step 5: Verify the solution

Substitute \(x = 8\) back into the original equation: \(8 + 7 = 15\) ✓

\(x = 8\)
Final answer:

\(x = 8\)

Applied rules:

• Inverse Operation: Subtraction undoes addition

• Balanced Equation: Whatever you do to one side, do to the other

• Variable Isolation: Get the variable alone on one side

2 Subtraction Equation
Exercise 2
Solve for \(y\):
\(y - 9 = 4\)
Definition:

One-Step Subtraction Equation: An equation where the variable is decreased by a constant

Original Equation
\(y - 9 = 4\)
Add 9 to both sides
\(y - 9 + 9 = 4 + 9\)
Simplify
\(y = 13\)
Step 1: Identify the operation

The variable \(y\) is being decreased by 9

Step 2: Apply the inverse operation

To undo subtracting 9, add 9 to both sides: \(y - 9 + 9 = 4 + 9\)

Step 3: Simplify

Left side: \(y - 9 + 9 = y\)

Right side: \(4 + 9 = 13\)

Step 4: Write the solution

\(y = 13\)

Step 5: Verify the solution

Substitute \(y = 13\) back into the original equation: \(13 - 9 = 4\) ✓

\(y = 13\)
Final answer:

\(y = 13\)

Applied rules:

• Inverse Operation: Addition undoes subtraction

• Balanced Equation: Whatever you do to one side, do to the other

• Variable Isolation: Get the variable alone on one side

3 Multiplication Equation
Exercise 3
Solve for \(z\):
\(6z = 42\)
Definition:

One-Step Multiplication Equation: An equation where the variable is multiplied by a constant

Original Equation
\(6z = 42\)
Divide both sides by 6
\(\frac{6z}{6} = \frac{42}{6}\)
Simplify
\(z = 7\)
Step 1: Identify the operation

The variable \(z\) is being multiplied by 6

Step 2: Apply the inverse operation

To undo multiplying by 6, divide both sides by 6: \(\frac{6z}{6} = \frac{42}{6}\)

Step 3: Simplify

Left side: \(\frac{6z}{6} = z\)

Right side: \(\frac{42}{6} = 7\)

Step 4: Write the solution

\(z = 7\)

Step 5: Verify the solution

Substitute \(z = 7\) back into the original equation: \(6 \times 7 = 42\) ✓

\(z = 7\)
Final answer:

\(z = 7\)

Applied rules:

• Inverse Operation: Division undoes multiplication

• Balanced Equation: Whatever you do to one side, do to the other

• Variable Isolation: Get the variable alone on one side

One-Step Equations Rules and Methods
\(x + a = b \Rightarrow x = b - a\)
Addition Form
Addition
\(x + a = b \Rightarrow x = b - a\)
Subtract from both sides to undo addition
Subtraction
\(x - a = b \Rightarrow x = b + a\)
Add to both sides to undo subtraction
Multiplication
\(ax = b \Rightarrow x = \frac{b}{a}\)
Divide both sides to undo multiplication
Division
\(\frac{x}{a} = b \Rightarrow x = ab\)
Multiply both sides to undo division
Inverse Operations
Addition ↔ Subtraction
Multiplication ↔ Division
Balanced Equation
Whatever you do to one side, do to the other
Maintain equality
Key Concepts: One-step equations require exactly one operation to isolate the variable. Always perform the same operation on both sides to maintain balance.
Verification: Always substitute your solution back into the original equation to check accuracy.
Tip 1: Identify what operation is being performed on the variable, then apply its inverse.
Tip 2: Always check your answer by substituting it back into the original equation.
Tip 3: Remember that division by zero is undefined, so check that your divisor is not zero.
Solution: Exercises 4 to 5
4 Division Equation
Exercise 4
Solve for \(w\):
\(\frac{w}{5} = 8\)
Definition:

One-Step Division Equation: An equation where the variable is divided by a constant

Original Equation
\(\frac{w}{5} = 8\)
Multiply both sides by 5
\(\frac{w}{5} \times 5 = 8 \times 5\)
Simplify
\(w = 40\)
Step 1: Identify the operation

The variable \(w\) is being divided by 5

Step 2: Apply the inverse operation

To undo dividing by 5, multiply both sides by 5: \(\frac{w}{5} \times 5 = 8 \times 5\)

Step 3: Simplify

Left side: \(\frac{w}{5} \times 5 = w\)

Right side: \(8 \times 5 = 40\)

Step 4: Write the solution

\(w = 40\)

Step 5: Verify the solution

Substitute \(w = 40\) back into the original equation: \(\frac{40}{5} = 8\) ✓

\(w = 40\)
Final answer:

\(w = 40\)

Applied rules:

• Inverse Operation: Multiplication undoes division

• Balanced Equation: Whatever you do to one side, do to the other

• Variable Isolation: Get the variable alone on one side

5 Mixed Operations with Fractions
Exercise 5
Solve for \(t\):
\(\frac{3}{4}t = 12\)
Definition:

One-Step Fraction Equation: An equation where the variable is multiplied by a fraction

Original Equation
\(\frac{3}{4}t = 12\)
Multiply both sides by reciprocal
\(\frac{3}{4}t \times \frac{4}{3} = 12 \times \frac{4}{3}\)
Simplify
\(t = \frac{48}{3} = 16\)
Step 1: Identify the operation

The variable \(t\) is being multiplied by \(\frac{3}{4}\)

Step 2: Apply the inverse operation

To undo multiplying by \(\frac{3}{4}\), multiply both sides by its reciprocal \(\frac{4}{3}\)

Step 3: Simplify the left side

\(\frac{3}{4}t \times \frac{4}{3} = \frac{3 \times 4}{4 \times 3}t = \frac{12}{12}t = t\)

Step 4: Simplify the right side

\(12 \times \frac{4}{3} = \frac{12 \times 4}{3} = \frac{48}{3} = 16\)

Step 5: Write the solution

\(t = 16\)

Step 6: Verify the solution

Substitute \(t = 16\) back into the original equation: \(\frac{3}{4} \times 16 = \frac{48}{4} = 12\) ✓

\(t = 16\)
Final answer:

\(t = 16\)

Applied rules:

• Inverse Operation: Multiply by reciprocal to undo fraction multiplication

• Balanced Equation: Whatever you do to one side, do to the other

• Variable Isolation: Get the variable alone on one side

Complete Guide: One-Step Equations, Rules, Methods, and Applications
\(ax + b = c \Rightarrow x = \frac{c - b}{a}\)
General Form
Key definitions:

Equation: A mathematical statement showing that two expressions are equal

Variable: A symbol representing an unknown value

One-Step Equation: An equation requiring only one operation to solve

Inverse Operations: Operations that undo each other (addition/subtraction, multiplication/division)

Complete methodology:
  1. Identify the operation: Determine what operation is being performed on the variable
  2. Select the inverse: Choose the operation that will undo the original operation
  3. Apply to both sides: Perform the inverse operation on both sides of the equation
  4. Simplify: Reduce both sides to isolate the variable
  5. Verify: Substitute the solution back into the original equation
Tip 1: Always perform the same operation on both sides to maintain equality.
Tip 2: When dividing by a fraction, multiply by its reciprocal.
Tip 3: Always check your solution by substituting it back into the original equation.
Tip 4: Be careful with signs when working with negative numbers.
Common errors: Forgetting to apply operations to both sides, incorrectly applying inverse operations, making sign errors with negative numbers.
Real-World Applications: One-step equations are used in everyday situations like calculating costs, determining distances, and solving rate problems.
Essential rules to memorize:

• Addition: \(x + a = b \Rightarrow x = b - a\)

• Subtraction: \(x - a = b \Rightarrow x = b + a\)

• Multiplication: \(ax = b \Rightarrow x = \frac{b}{a}\)

• Division: \(\frac{x}{a} = b \Rightarrow x = ab\)

• Balanced Equation: Perform same operation on both sides

• Verification: Always check your solution

Exercise with Visualization: One-Step Equations Solutions
Exercise 6: Equation Solutions Comparison
Consider the following equations and their solutions:
\(f_1(x) = x + 7 = 15 \Rightarrow x = 8\)
\(f_2(y) = y - 9 = 4 \Rightarrow y = 13\)
\(f_3(z) = 6z = 42 \Rightarrow z = 7\)

Analysis: The chart shows how different one-step equations have different solutions.

  • \(x + 7 = 15\) (solution: \(x = 8\))
  • \(y - 9 = 4\) (solution: \(y = 13\))
  • \(6z = 42\) (solution: \(z = 7\))

Questions & Answers

Question: Why do I need to do the same thing to both sides of the equation? Can't I just move numbers around?

Answer: Think of an equation like a balanced scale - both sides must remain equal.

If you have \(x + 5 = 10\), imagine a scale with \(x + 5\) on one side and \(10\) on the other. They balance perfectly.

If you subtract 5 from only one side, the scale becomes unbalanced:

  • Left side: \(x + 5 - 5 = x\)
  • Right side: \(10\) (unchanged)
  • Now: \(x = 10\) (which is wrong!)

But if you subtract 5 from both sides:

  • Left side: \(x + 5 - 5 = x\)
  • Right side: \(10 - 5 = 5\)
  • Now: \(x = 5\) (which is correct!)

The equals sign means both sides are equal, so whatever you do to one side, you must do to the other to maintain balance.

Question: How do I know which inverse operation to use? Like when do I add vs subtract?

Answer: Look at what's happening to your variable and do the opposite:

Operation on Variable → Inverse Operation to Undo:

  • \(x + 7 = 15\) → The variable has 7 added to it, so subtract 7 from both sides
  • \(y - 9 = 4\) → The variable has 9 subtracted from it, so add 9 to both sides
  • \(6z = 42\) → The variable is multiplied by 6, so divide both sides by 6
  • \(\frac{w}{5} = 8\) → The variable is divided by 5, so multiply both sides by 5

Think of it like unwrapping a gift - you reverse the steps that were done to the variable.

If someone put a hat on you (+hat), you take off the hat (-hat).

If someone gave you 3 apples (×3), you divide by 3 to get back to 1 apple.

Question: What if I have a negative number in the equation? Like \(x + (-3) = 7\)?

Answer: You can simplify first or work with the negative directly:

Method 1 (Simplify first):

  • \(x + (-3) = 7\) becomes \(x - 3 = 7\)
  • To undo subtracting 3, add 3 to both sides: \(x - 3 + 3 = 7 + 3\)
  • Result: \(x = 10\)

Method 2 (Work with negative):

  • \(x + (-3) = 7\)
  • To undo adding \(-3\), subtract \(-3\) from both sides: \(x + (-3) - (-3) = 7 - (-3)\)
  • Simplify: \(x = 7 + 3 = 10\)

The first method is generally easier - simplify the equation first, then solve.

Remember: Adding a negative number is the same as subtracting a positive number.

Question: How do I solve equations with fractions like \(\frac{2}{3}x = 8\)? Why do I multiply by the reciprocal?

Answer: To undo multiplication by a fraction, you multiply by its reciprocal:

For \(\frac{2}{3}x = 8\):

  1. The variable \(x\) is multiplied by \(\frac{2}{3}\)
  2. The inverse of \(\frac{2}{3}\) is \(\frac{3}{2}\) (flip the numerator and denominator)
  3. Multiply both sides by \(\frac{3}{2}\): \(\frac{2}{3}x \times \frac{3}{2} = 8 \times \frac{3}{2}\)
  4. Left side: \(\frac{2}{3} \times \frac{3}{2} \times x = \frac{6}{6} \times x = 1 \times x = x\)
  5. Right side: \(8 \times \frac{3}{2} = \frac{24}{2} = 12\)
  6. Result: \(x = 12\)

Why does this work? Because \(\frac{a}{b} \times \frac{b}{a} = \frac{ab}{ba} = 1\).

Multiplying by the reciprocal turns the coefficient into 1, leaving just the variable.

Question: How do I know if my answer is correct? What if I made a mistake?

Answer: Always substitute your solution back into the ORIGINAL equation to verify:

Example: Solve \(x + 7 = 15\), you get \(x = 8\)

  1. Take the original equation: \(x + 7 = 15\)
  2. Replace \(x\) with your answer (8): \(8 + 7 = 15\)
  3. Calculate: \(15 = 15\) ✓ (True! Your answer is correct)

If you get something like \(14 = 15\), then your answer is wrong and you need to recheck your work.

This verification step is crucial - it catches errors and confirms your solution is correct.

Example of error: If you thought \(x = 9\) for \(x + 7 = 15\):

  • Substitute: \(9 + 7 = 15\)
  • Calculate: \(16 = 15\) ✗ (False! So \(x = 9\) is wrong)

Always verify your answers!