1. Which x Satisfies an Equation?

Definition: To determine if a value satisfies an equation, substitute the value for the variable and check if both sides are equal.

Steps to Check:

1. Substitute the given value for the variable
2. Simplify both sides of the equation
3. Check if left side equals right side
4. If equal, the value is a solution; if not, it isn't

Examples:

Example 1: Does \( x = 5 \) satisfy the equation \( 3x + 4 = 19 \)?

Substitute: \( 3(5) + 4 = 19 \)

Simplify: \( 15 + 4 = 19 \)

\( 19 = 19 \) ✓ Yes, \( x = 5 \) is a solution!

Example 2: Does \( x = 3 \) satisfy the equation \( 2x - 7 = 5 \)?

Substitute: \( 2(3) - 7 = 5 \)

Simplify: \( 6 - 7 = 5 \)

\( -1 \neq 5 \) ✗ No, \( x = 3 \) is NOT a solution!

2. Write an Equation from Words

Key Translation Words:

Steps:

1. Identify the unknown (assign a variable)
2. Translate key words to operations
3. Write the left side of the equation
4. Write the right side of the equation

Examples:

Example 1: "Five more than a number is 12"

Let \( x \) = the number

Equation: \( x + 5 = 12 \)

Example 2: "Three times a number, decreased by 7, equals 20"

Let \( n \) = the number

Equation: \( 3n - 7 = 20 \)

Example 3: "The quotient of a number and 4 is 9"

Let \( x \) = the number

Equation: \( \frac{x}{4} = 9 \)

3. Model and Solve Equations Using Algebra Tiles

Algebra Tiles Representation:

• Large rectangle: represents \( x \) (variable)
• Small square: represents 1 (unit or constant)
• Positive tiles: shown in one color (e.g., green)
• Negative tiles: shown in opposite color (e.g., red)

Key Concept:

Zero Pair: One positive tile and one negative tile cancel each other out (equal zero)

Steps to Solve:

1. Model the equation with tiles
2. Remove zero pairs
3. Isolate the variable tiles on one side
4. Divide the constant tiles equally by the number of variable tiles

Example:

Solve: \( x + 3 = 7 \)

Model: 1 \( x \)-tile + 3 unit tiles = 7 unit tiles

Remove 3 unit tiles from both sides

Result: 1 \( x \)-tile = 4 unit tiles

Solution: \( x = 4 \)

4. Properties of Equality

Definition: Properties that maintain equality when the same operation is performed on both sides of an equation.

Addition Property of Equality:

If \( a = b \), then \( a + c = b + c \)

Example: If \( x - 5 = 10 \), add 5 to both sides: \( x = 15 \)

Subtraction Property of Equality:

If \( a = b \), then \( a - c = b - c \)

Example: If \( x + 7 = 12 \), subtract 7 from both sides: \( x = 5 \)

Multiplication Property of Equality:

If \( a = b \), then \( a \times c = b \times c \) (where \( c \neq 0 \))

Example: If \( \frac{x}{3} = 4 \), multiply both sides by 3: \( x = 12 \)

Division Property of Equality:

If \( a = b \), then \( \frac{a}{c} = \frac{b}{c} \) (where \( c \neq 0 \))

Example: If \( 4x = 20 \), divide both sides by 4: \( x = 5 \)

Reflexive Property:

\( a = a \) (Any number equals itself)

Symmetric Property:

If \( a = b \), then \( b = a \)

Transitive Property:

If \( a = b \) and \( b = c \), then \( a = c \)

Substitution Property:

If \( a = b \), then \( a \) can replace \( b \) in any expression

5. Solve One-Step Equations

Definition: An equation that requires only ONE operation to solve for the variable.

General Steps:

1. Identify the operation acting on the variable
2. Use the inverse operation on both sides
3. Simplify to find the solution

Type 1: Addition Equations

Form: \( x + a = b \)

Solution: Subtract \( a \) from both sides: \( x = b - a \)

Example: \( x + 9 = 15 \)

\( x + 9 - 9 = 15 - 9 \)

\( x = 6 \)

Type 2: Subtraction Equations

Form: \( x - a = b \)

Solution: Add \( a \) to both sides: \( x = b + a \)

Example: \( x - 7 = 12 \)

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

\( x = 19 \)

Type 3: Multiplication Equations

Form: \( ax = b \)

Solution: Divide both sides by \( a \): \( x = \frac{b}{a} \)

Example: \( 5x = 35 \)

\( \frac{5x}{5} = \frac{35}{5} \)

\( x = 7 \)

Type 4: Division Equations

Form: \( \frac{x}{a} = b \)

Solution: Multiply both sides by \( a \): \( x = ab \)

Example: \( \frac{x}{4} = 8 \)

\( 4 \cdot \frac{x}{4} = 4 \cdot 8 \)

\( x = 32 \)

6. Solve Two-Step Equations

Definition: An equation that requires TWO operations to solve for the variable.

General Form:

\( ax + b = c \) or \( \frac{x}{a} + b = c \)

Steps to Solve:

1. Step 1: Use addition or subtraction to isolate the term with the variable
2. Step 2: Use multiplication or division to solve for the variable

Examples:

Example 1: Solve \( 3x + 7 = 22 \)

Step 1: Subtract 7 from both sides

\( 3x + 7 - 7 = 22 - 7 \)

\( 3x = 15 \)

Step 2: Divide both sides by 3

\( x = 5 \)

Example 2: Solve \( \frac{x}{5} - 3 = 4 \)

Step 1: Add 3 to both sides

\( \frac{x}{5} - 3 + 3 = 4 + 3 \)

\( \frac{x}{5} = 7 \)

Step 2: Multiply both sides by 5

\( x = 35 \)

Example 3: Solve \( -2y + 9 = 3 \)

Subtract 9: \( -2y = -6 \)

Divide by -2: \( y = 3 \)

7. Solve Equations Involving Like Terms

Steps:

1. Combine like terms on each side of the equation
2. Use inverse operations to isolate the variable
3. Solve for the variable

Examples:

Example 1: Solve \( 5x + 2x - 3 = 18 \)

Step 1: Combine like terms: \( 7x - 3 = 18 \)

Step 2: Add 3: \( 7x = 21 \)

Step 3: Divide by 7: \( x = 3 \)

Example 2: Solve \( 8y - 3y + 5 = 20 \)

Combine: \( 5y + 5 = 20 \)

Subtract 5: \( 5y = 15 \)

Divide by 5: \( y = 3 \)

8. Solve Equations with Variables on Both Sides

Steps:

1. Combine like terms on each side (if needed)
2. Move all variable terms to one side (usually the left)
3. Move all constant terms to the other side
4. Solve for the variable

Examples:

Example 1: Solve \( 5x + 3 = 2x + 12 \)

Step 1: Subtract \( 2x \) from both sides

\( 5x - 2x + 3 = 2x - 2x + 12 \)

\( 3x + 3 = 12 \)

Step 2: Subtract 3 from both sides

\( 3x = 9 \)

Step 3: Divide by 3

\( x = 3 \)

Example 2: Solve \( 7y - 4 = 3y + 8 \)

Subtract \( 3y \): \( 4y - 4 = 8 \)

Add 4: \( 4y = 12 \)

Divide by 4: \( y = 3 \)

With Fractional Coefficients:

Example 3: Solve \( \frac{2x}{3} + 5 = \frac{x}{2} + 8 \)

Method: Multiply all terms by the LCD (6)

\( 6 \cdot \frac{2x}{3} + 6 \cdot 5 = 6 \cdot \frac{x}{2} + 6 \cdot 8 \)

\( 4x + 30 = 3x + 48 \)

Subtract \( 3x \): \( x + 30 = 48 \)

Subtract 30: \( x = 18 \)

9. Solve Equations Using the Distributive Property

Steps:

1. Apply the distributive property: \( a(b + c) = ab + ac \)
2. Combine like terms
3. Use inverse operations to solve

Examples:

Example 1: Solve \( 3(x + 4) = 21 \)

Step 1: Distribute: \( 3x + 12 = 21 \)

Step 2: Subtract 12: \( 3x = 9 \)

Step 3: Divide by 3: \( x = 3 \)

Example 2: Solve \( 5(2y - 3) + 7 = 32 \)

Distribute: \( 10y - 15 + 7 = 32 \)

Combine: \( 10y - 8 = 32 \)

Add 8: \( 10y = 40 \)

Divide by 10: \( y = 4 \)

Example 3: Solve \( 2(x - 5) = 3(x + 1) \)

Distribute both sides: \( 2x - 10 = 3x + 3 \)

Subtract \( 2x \): \( -10 = x + 3 \)

Subtract 3: \( x = -13 \)

10. Solve Multi-Step Equations

Definition: Equations requiring three or more steps to solve.

General Strategy:

1. Simplify each side (distribute, combine like terms)
2. Move variable terms to one side
3. Move constant terms to the other side
4. Solve for the variable

Example:

Solve: \( 4(2x - 3) + 5x = 3(x + 7) - 8 \)

Step 1: Distribute

\( 8x - 12 + 5x = 3x + 21 - 8 \)

Step 2: Combine like terms on each side

\( 13x - 12 = 3x + 13 \)

Step 3: Subtract \( 3x \) from both sides

\( 10x - 12 = 13 \)

Step 4: Add 12 to both sides

\( 10x = 25 \)

Step 5: Divide by 10

\( x = \frac{25}{10} = \frac{5}{2} = 2.5 \)

11. Find the Number of Solutions

Linear equations can have:

One Solution (Unique Solution):

When solving results in a specific value for the variable

Example: \( 2x + 3 = 7 \) → \( x = 2 \)

Characteristic: Different coefficients for the variable on each side

No Solution:

When solving results in a false statement (like \( 0 = 5 \) or \( 3 = 7 \))

Example: \( 2x + 3 = 2x + 7 \)

Subtract \( 2x \): \( 3 = 7 \) (False!) → No solution

Characteristic: Same coefficients for variables, different constants

Notation: ∅ or "no solution"

Infinitely Many Solutions:

When solving results in a true statement (like \( 0 = 0 \) or \( 5 = 5 \))

Example: \( 3x + 6 = 3(x + 2) \)

\( 3x + 6 = 3x + 6 \)

Subtract \( 3x \): \( 6 = 6 \) (True!) → Infinite solutions

Characteristic: Both sides are identical (equivalent expressions)

Notation: "All real numbers" or \( x \in \mathbb{R} \)

Summary Table:

12. Create Equations with No Solutions or Infinitely Many Solutions

To Create NO SOLUTION:

Make the variable terms equal on both sides, but constants different

Template: \( ax + b = ax + c \) where \( b \neq c \)

Example: \( 5x + 3 = 5x + 8 \) (No solution)

Example: \( 2(x + 4) = 2x + 10 \) → \( 2x + 8 = 2x + 10 \) (No solution)

To Create INFINITE SOLUTIONS:

Make both sides identical (equivalent expressions)

Template: \( ax + b = ax + b \)

Example: \( 4x - 7 = 4x - 7 \) (Infinite solutions)

Example: \( 3(x + 2) = 3x + 6 \) → \( 3x + 6 = 3x + 6 \) (Infinite solutions)

Practice Problems:

Problem 1: Fill in the blank to create no solution: \( 7x + 5 = 7x + ___ \)

Answer: Any number except 5 (e.g., 8, 12, -3, etc.)

Problem 2: Fill in the blank to create infinite solutions: \( 3(2x - 4) = 6x - ___ \)

Answer: 12 (because \( 3(2x - 4) = 6x - 12 \))

Problem 3: Determine the value of \( k \) that makes \( 4x + k = 4x + 9 \) have no solution.

Answer: Any value except 9 (e.g., \( k = 3 \), \( k = 0 \), etc.)

13. Word Problems with Equations

General Steps:

1. Read the problem carefully
2. Identify what you're looking for (assign a variable)
3. Write an equation from the given information
4. Solve the equation
5. Check if the answer makes sense in context
6. Write a complete answer with units

Example Problems:

Example 1: Sarah has $45. She wants to buy a book for $12 and some pencils that cost $3 each. How many pencils can she buy?

Let \( p \) = number of pencils

Equation: \( 12 + 3p = 45 \)

Subtract 12: \( 3p = 33 \)

Divide by 3: \( p = 11 \)

Answer: Sarah can buy 11 pencils.

Example 2: The sum of three consecutive integers is 48. Find the integers.

Let \( n \) = first integer

Then \( n + 1 \) and \( n + 2 \) are the next two

Equation: \( n + (n + 1) + (n + 2) = 48 \)

Simplify: \( 3n + 3 = 48 \)

Subtract 3: \( 3n = 45 \)

Divide by 3: \( n = 15 \)

Answer: The integers are 15, 16, and 17.

Example 3: A rectangle's length is 5 cm more than twice its width. If the perimeter is 46 cm, find the dimensions.

Let \( w \) = width

Then \( l = 2w + 5 \)

Perimeter: \( 2l + 2w = 46 \)

Substitute: \( 2(2w + 5) + 2w = 46 \)

\( 4w + 10 + 2w = 46 \)

\( 6w + 10 = 46 \)

\( 6w = 36 \) → \( w = 6 \) cm

\( l = 2(6) + 5 = 17 \) cm

Answer: Width = 6 cm, Length = 17 cm

Quick Reference: Equation Solving

💡 Key Tips for Solving Equations

• ✓ Always do the same operation to both sides to maintain equality
• ✓ Use inverse operations: Addition ↔ Subtraction, Multiplication ↔ Division
• ✓ Simplify first: Distribute and combine like terms before isolating the variable
• ✓ Move variables to one side: Usually move to the side with the larger coefficient
• ✓ Work with fractions: Multiply by LCD to clear denominators
• ✓ Check your solution: Substitute back into the original equation
• ✓ Watch for special cases: No solution (false statement) or infinite solutions (true statement)
• ✓ Order matters: Undo operations in reverse order (PEMDAS backwards)
• ✓ Be careful with negatives: Especially when distributing or moving terms
• ✓ Show your work: Write each step clearly to avoid errors