Solve Linear Equations - Ninth Grade Math
1. Write Variable Equations
Equation: A mathematical statement that two expressions are equal, connected by an equals sign (=)
Variable Equation: An equation that contains at least one variable
Solution: The value of the variable that makes the equation true
Variable Equation: An equation that contains at least one variable
Solution: The value of the variable that makes the equation true
Translating Word Problems to Equations
| Phrase | Equation |
|---|---|
| "A number increased by 7 is 15" | $x + 7 = 15$ |
| "5 less than a number equals 12" | $x - 5 = 12$ |
| "The product of 3 and a number is 21" | $3x = 21$ |
| "A number divided by 4 equals 8" | $\frac{x}{4} = 8$ |
| "Twice a number, decreased by 5, is 11" | $2x - 5 = 11$ |
| "The sum of a number and 10 is 3 times the number" | $x + 10 = 3x$ |
Practice Examples:
1. "Seven more than twice a number is 23"
Equation: $2x + 7 = 23$
2. "The difference between a number and 8 is 15"
Equation: $x - 8 = 15$
3. "One-third of a number equals 12"
Equation: $\frac{x}{3} = 12$ or $\frac{1}{3}x = 12$
1. "Seven more than twice a number is 23"
Equation: $2x + 7 = 23$
2. "The difference between a number and 8 is 15"
Equation: $x - 8 = 15$
3. "One-third of a number equals 12"
Equation: $\frac{x}{3} = 12$ or $\frac{1}{3}x = 12$
2. Does x Satisfy the Equation?
Satisfy an Equation: When a value substituted for the variable makes the equation true
Check a Solution: Substitute the value and verify if LHS = RHS
Check a Solution: Substitute the value and verify if LHS = RHS
Steps to Check if a Value Satisfies an Equation:
Step 1: Substitute the given value for the variable
Step 2: Simplify both sides of the equation
Step 3: Check if Left-Hand Side (LHS) = Right-Hand Side (RHS)
Step 4: If LHS = RHS, the value satisfies the equation ✓
If LHS ≠ RHS, the value does NOT satisfy the equation ✗
Step 1: Substitute the given value for the variable
Step 2: Simplify both sides of the equation
Step 3: Check if Left-Hand Side (LHS) = Right-Hand Side (RHS)
Step 4: If LHS = RHS, the value satisfies the equation ✓
If LHS ≠ RHS, the value does NOT satisfy the equation ✗
Example 1: Does $x = 5$ satisfy the equation $2x + 3 = 13$?
Substitute: $2(5) + 3 = 13$
Simplify: $10 + 3 = 13$
Result: $13 = 13$ ✓
Yes, $x = 5$ satisfies the equation
Substitute: $2(5) + 3 = 13$
Simplify: $10 + 3 = 13$
Result: $13 = 13$ ✓
Yes, $x = 5$ satisfies the equation
Example 2: Does $x = 4$ satisfy the equation $3x - 5 = 10$?
Substitute: $3(4) - 5 = 10$
Simplify: $12 - 5 = 10$
Result: $7 \neq 10$ ✗
No, $x = 4$ does NOT satisfy the equation
Substitute: $3(4) - 5 = 10$
Simplify: $12 - 5 = 10$
Result: $7 \neq 10$ ✗
No, $x = 4$ does NOT satisfy the equation
Example 3: Does $x = -2$ satisfy the equation $5x + 8 = -2$?
Substitute: $5(-2) + 8 = -2$
Simplify: $-10 + 8 = -2$
Result: $-2 = -2$ ✓
Yes, $x = -2$ satisfies the equation
Substitute: $5(-2) + 8 = -2$
Simplify: $-10 + 8 = -2$
Result: $-2 = -2$ ✓
Yes, $x = -2$ satisfies the equation
3. Which x Satisfies an Equation?
Method: Test each given value by substitution to find which one makes the equation true
Example: Which value satisfies $4x - 7 = 9$? Test: $x = 2$, $x = 3$, $x = 4$, $x = 5$
Test $x = 2$: $4(2) - 7 = 8 - 7 = 1 \neq 9$ ✗
Test $x = 3$: $4(3) - 7 = 12 - 7 = 5 \neq 9$ ✗
Test $x = 4$: $4(4) - 7 = 16 - 7 = 9 = 9$ ✓
Test $x = 5$: Not needed (already found)
Answer: $x = 4$ satisfies the equation
Test $x = 2$: $4(2) - 7 = 8 - 7 = 1 \neq 9$ ✗
Test $x = 3$: $4(3) - 7 = 12 - 7 = 5 \neq 9$ ✗
Test $x = 4$: $4(4) - 7 = 16 - 7 = 9 = 9$ ✓
Test $x = 5$: Not needed (already found)
Answer: $x = 4$ satisfies the equation
Tip: Once you find a value that works, you can stop testing (linear equations have only one solution)
4. Properties of Equality
Properties of Equality: Rules that allow you to perform the same operation on both sides of an equation without changing the solution
Addition Property of Equality
If $a = b$, then $a + c = b + c$
You can add the same number to both sides of an equation
You can add the same number to both sides of an equation
Example: If $x - 5 = 10$, add 5 to both sides
$x - 5 + 5 = 10 + 5$
$x = 15$
$x - 5 + 5 = 10 + 5$
$x = 15$
Subtraction Property of Equality
If $a = b$, then $a - c = b - c$
You can subtract the same number from both sides of an equation
You can subtract the same number from both sides of an equation
Example: If $x + 8 = 12$, subtract 8 from both sides
$x + 8 - 8 = 12 - 8$
$x = 4$
$x + 8 - 8 = 12 - 8$
$x = 4$
Multiplication Property of Equality
If $a = b$, then $a \cdot c = b \cdot c$
You can multiply both sides of an equation by the same non-zero number
You can multiply both sides of an equation by the same non-zero number
Example: If $\frac{x}{3} = 6$, multiply both sides by 3
$3 \cdot \frac{x}{3} = 3 \cdot 6$
$x = 18$
$3 \cdot \frac{x}{3} = 3 \cdot 6$
$x = 18$
Division Property of Equality
If $a = b$, then $\frac{a}{c} = \frac{b}{c}$ (where $c \neq 0$)
You can divide both sides of an equation by the same non-zero number
You can divide both sides of an equation by the same non-zero number
Example: If $5x = 20$, divide both sides by 5
$\frac{5x}{5} = \frac{20}{5}$
$x = 4$
$\frac{5x}{5} = \frac{20}{5}$
$x = 4$
Other Properties of Equality
| Property | Statement | Example |
|---|---|---|
| Reflexive Property | $a = a$ | $5 = 5$, $x = x$ |
| Symmetric Property | If $a = b$, then $b = a$ | If $x = 7$, then $7 = x$ |
| Transitive Property | If $a = b$ and $b = c$, then $a = c$ | If $x = 5$ and $5 = y$, then $x = y$ |
| Substitution Property | If $a = b$, then $a$ can replace $b$ | If $x = 3$, then $2x = 2(3) = 6$ |
5. Identify Equivalent Equations
Equivalent Equations: Equations that have the same solution(s)
Example: $x + 3 = 7$ and $x = 4$ are equivalent (both have solution $x = 4$)
Example: $x + 3 = 7$ and $x = 4$ are equivalent (both have solution $x = 4$)
How to Check if Equations are Equivalent:
• Solve both equations
• If they have the same solution, they are equivalent
• OR: Use properties of equality to transform one equation into the other
• Solve both equations
• If they have the same solution, they are equivalent
• OR: Use properties of equality to transform one equation into the other
Example 1: Are $2x + 6 = 14$ and $x + 3 = 7$ equivalent?
Method 1 - Solve both:
First equation: $2x + 6 = 14$ → $2x = 8$ → $x = 4$
Second equation: $x + 3 = 7$ → $x = 4$
Same solution ✓ → They are equivalent
Method 2 - Transform:
Divide first equation by 2: $\frac{2x + 6}{2} = \frac{14}{2}$ → $x + 3 = 7$ ✓
Method 1 - Solve both:
First equation: $2x + 6 = 14$ → $2x = 8$ → $x = 4$
Second equation: $x + 3 = 7$ → $x = 4$
Same solution ✓ → They are equivalent
Method 2 - Transform:
Divide first equation by 2: $\frac{2x + 6}{2} = \frac{14}{2}$ → $x + 3 = 7$ ✓
Example 2: Are $3x - 5 = 10$ and $3x = 15$ equivalent?
Transform first equation: Add 5 to both sides
$3x - 5 + 5 = 10 + 5$
$3x = 15$ ✓
Yes, they are equivalent
Transform first equation: Add 5 to both sides
$3x - 5 + 5 = 10 + 5$
$3x = 15$ ✓
Yes, they are equivalent
6. Model and Solve Linear Equations Using Algebra Tiles
Algebra Tiles: Visual manipulatives used to represent equations
• Large square = $x^2$ (not used in linear equations)
• Rectangle = $x$ (variable)
• Small square = $1$ (unit/constant)
• Different colors represent positive and negative values
• Large square = $x^2$ (not used in linear equations)
• Rectangle = $x$ (variable)
• Small square = $1$ (unit/constant)
• Different colors represent positive and negative values
Key Concepts for Algebra Tiles:
• One positive tile + one negative tile = zero pair (cancel out)
• Keep equation balanced: same operations on both sides
• Goal: Isolate $x$ tiles on one side
• One positive tile + one negative tile = zero pair (cancel out)
• Keep equation balanced: same operations on both sides
• Goal: Isolate $x$ tiles on one side
Example: Solve $x + 3 = 7$ using algebra tiles
Visual Representation:
Left side: 1 rectangle ($x$) + 3 small squares (units)
Right side: 7 small squares (units)
Solution Process:
Remove 3 unit tiles from both sides (keeping balance)
Left side: 1 rectangle ($x$)
Right side: 4 small squares
Answer: $x = 4$
Visual Representation:
Left side: 1 rectangle ($x$) + 3 small squares (units)
Right side: 7 small squares (units)
Solution Process:
Remove 3 unit tiles from both sides (keeping balance)
Left side: 1 rectangle ($x$)
Right side: 4 small squares
Answer: $x = 4$
Principle: Whatever you do to one side, do to the other side to maintain balance
7. Solve One-Step Linear Equations
One-Step Equation: An equation that requires only ONE operation to solve
Goal: Isolate the variable using the inverse operation
Goal: Isolate the variable using the inverse operation
Types of One-Step Equations
Type 1: Addition Equation ($x + a = b$)
Inverse Operation: Subtract $a$ from both sides
Solution: $x = b - a$
Inverse Operation: Subtract $a$ from both sides
Solution: $x = b - a$
Example: $x + 7 = 15$
Subtract 7: $x + 7 - 7 = 15 - 7$
Solution: $x = 8$
Subtract 7: $x + 7 - 7 = 15 - 7$
Solution: $x = 8$
Type 2: Subtraction Equation ($x - a = b$)
Inverse Operation: Add $a$ to both sides
Solution: $x = b + a$
Inverse Operation: Add $a$ to both sides
Solution: $x = b + a$
Example: $x - 9 = 4$
Add 9: $x - 9 + 9 = 4 + 9$
Solution: $x = 13$
Add 9: $x - 9 + 9 = 4 + 9$
Solution: $x = 13$
Type 3: Multiplication Equation ($ax = b$)
Inverse Operation: Divide both sides by $a$
Solution: $x = \frac{b}{a}$
Inverse Operation: Divide both sides by $a$
Solution: $x = \frac{b}{a}$
Example: $6x = 42$
Divide by 6: $\frac{6x}{6} = \frac{42}{6}$
Solution: $x = 7$
Divide by 6: $\frac{6x}{6} = \frac{42}{6}$
Solution: $x = 7$
Type 4: Division Equation ($\frac{x}{a} = b$)
Inverse Operation: Multiply both sides by $a$
Solution: $x = ab$
Inverse Operation: Multiply both sides by $a$
Solution: $x = ab$
Example: $\frac{x}{5} = 8$
Multiply by 5: $5 \cdot \frac{x}{5} = 5 \cdot 8$
Solution: $x = 40$
Multiply by 5: $5 \cdot \frac{x}{5} = 5 \cdot 8$
Solution: $x = 40$
| Equation Type | Example | Inverse Operation | Solution |
|---|---|---|---|
| $x + a = b$ | $x + 5 = 12$ | Subtract 5 | $x = 7$ |
| $x - a = b$ | $x - 3 = 10$ | Add 3 | $x = 13$ |
| $ax = b$ | $4x = 20$ | Divide by 4 | $x = 5$ |
| $\frac{x}{a} = b$ | $\frac{x}{2} = 9$ | Multiply by 2 | $x = 18$ |
8. Solve Two-Step Linear Equations
Two-Step Equation: An equation that requires TWO operations to solve
General Form: $ax + b = c$ or $\frac{x}{a} + b = c$
General Form: $ax + b = c$ or $\frac{x}{a} + b = c$
Steps to Solve Two-Step Equations:
Step 1: Undo addition or subtraction (isolate the term with variable)
Step 2: Undo multiplication or division (isolate the variable)
Remember: Reverse order of operations (undo addition/subtraction FIRST)
Step 1: Undo addition or subtraction (isolate the term with variable)
Step 2: Undo multiplication or division (isolate the variable)
Remember: Reverse order of operations (undo addition/subtraction FIRST)
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
$\frac{3x}{3} = \frac{15}{3}$
Solution: $x = 5$
Check: $3(5) + 7 = 15 + 7 = 22$ ✓
Step 1: Subtract 7 from both sides
$3x + 7 - 7 = 22 - 7$
$3x = 15$
Step 2: Divide both sides by 3
$\frac{3x}{3} = \frac{15}{3}$
Solution: $x = 5$
Check: $3(5) + 7 = 15 + 7 = 22$ ✓
Example 2: Solve $\frac{x}{4} - 5 = 3$
Step 1: Add 5 to both sides
$\frac{x}{4} - 5 + 5 = 3 + 5$
$\frac{x}{4} = 8$
Step 2: Multiply both sides by 4
$4 \cdot \frac{x}{4} = 4 \cdot 8$
Solution: $x = 32$
Check: $\frac{32}{4} - 5 = 8 - 5 = 3$ ✓
Step 1: Add 5 to both sides
$\frac{x}{4} - 5 + 5 = 3 + 5$
$\frac{x}{4} = 8$
Step 2: Multiply both sides by 4
$4 \cdot \frac{x}{4} = 4 \cdot 8$
Solution: $x = 32$
Check: $\frac{32}{4} - 5 = 8 - 5 = 3$ ✓
Example 3: Solve $-2x + 9 = 3$
Step 1: Subtract 9 from both sides
$-2x + 9 - 9 = 3 - 9$
$-2x = -6$
Step 2: Divide both sides by -2
$\frac{-2x}{-2} = \frac{-6}{-2}$
Solution: $x = 3$
Check: $-2(3) + 9 = -6 + 9 = 3$ ✓
Step 1: Subtract 9 from both sides
$-2x + 9 - 9 = 3 - 9$
$-2x = -6$
Step 2: Divide both sides by -2
$\frac{-2x}{-2} = \frac{-6}{-2}$
Solution: $x = 3$
Check: $-2(3) + 9 = -6 + 9 = 3$ ✓
9. Solve One-Step and Two-Step Linear Equations: Word Problems
Steps for Solving Word Problems:
Step 1: Read the problem carefully
Step 2: Identify what you need to find (assign a variable)
Step 3: Write an equation based on the problem
Step 4: Solve the equation
Step 5: Check your answer in the context of the problem
Step 6: Write your answer in a complete sentence
Step 1: Read the problem carefully
Step 2: Identify what you need to find (assign a variable)
Step 3: Write an equation based on the problem
Step 4: Solve the equation
Step 5: Check your answer in the context of the problem
Step 6: Write your answer in a complete sentence
Example 1 (One-Step):
"Maria had some money. She spent $15 and has $42 left. How much money did she have originally?"
Let: $x$ = original amount of money
Equation: $x - 15 = 42$
Solve: $x = 42 + 15 = 57$
Answer: Maria originally had $57.
"Maria had some money. She spent $15 and has $42 left. How much money did she have originally?"
Let: $x$ = original amount of money
Equation: $x - 15 = 42$
Solve: $x = 42 + 15 = 57$
Answer: Maria originally had $57.
Example 2 (Two-Step):
"A taxi charges a $5 flat fee plus $2 per mile. If the total fare was $23, how many miles did you travel?"
Let: $x$ = number of miles
Equation: $2x + 5 = 23$
Solve:
Step 1: $2x = 23 - 5 = 18$
Step 2: $x = 18 \div 2 = 9$
Answer: You traveled 9 miles.
"A taxi charges a $5 flat fee plus $2 per mile. If the total fare was $23, how many miles did you travel?"
Let: $x$ = number of miles
Equation: $2x + 5 = 23$
Solve:
Step 1: $2x = 23 - 5 = 18$
Step 2: $x = 18 \div 2 = 9$
Answer: You traveled 9 miles.
Example 3 (Two-Step):
"The perimeter of a rectangle is 50 cm. The length is 15 cm. What is the width?"
(Note: Perimeter = $2l + 2w$)
Let: $w$ = width
Equation: $2(15) + 2w = 50$
Solve:
$30 + 2w = 50$
$2w = 20$
$w = 10$
Answer: The width is 10 cm.
"The perimeter of a rectangle is 50 cm. The length is 15 cm. What is the width?"
(Note: Perimeter = $2l + 2w$)
Let: $w$ = width
Equation: $2(15) + 2w = 50$
Solve:
$30 + 2w = 50$
$2w = 20$
$w = 10$
Answer: The width is 10 cm.
10. Solve Linear Equations with Variables on One Side
Multi-Step Equation: Equations requiring more than two steps, often involving distributive property and combining like terms
General Steps:
Step 1: Simplify each side (distribute, combine like terms)
Step 2: Use addition/subtraction to move constants
Step 3: Use multiplication/division to isolate variable
Step 4: Check your solution
Step 1: Simplify each side (distribute, combine like terms)
Step 2: Use addition/subtraction to move constants
Step 3: Use multiplication/division to isolate variable
Step 4: Check your solution
Example 1: Solve $3(x + 4) - 5 = 16$
Step 1: Distribute: $3x + 12 - 5 = 16$
Step 2: Combine like terms: $3x + 7 = 16$
Step 3: Subtract 7: $3x = 9$
Step 4: Divide by 3: $x = 3$
Solution: $x = 3$
Step 1: Distribute: $3x + 12 - 5 = 16$
Step 2: Combine like terms: $3x + 7 = 16$
Step 3: Subtract 7: $3x = 9$
Step 4: Divide by 3: $x = 3$
Solution: $x = 3$
Example 2: Solve $5x - 2(x - 3) = 15$
Step 1: Distribute: $5x - 2x + 6 = 15$
Step 2: Combine like terms: $3x + 6 = 15$
Step 3: Subtract 6: $3x = 9$
Step 4: Divide by 3: $x = 3$
Solution: $x = 3$
Step 1: Distribute: $5x - 2x + 6 = 15$
Step 2: Combine like terms: $3x + 6 = 15$
Step 3: Subtract 6: $3x = 9$
Step 4: Divide by 3: $x = 3$
Solution: $x = 3$
Example 3: Solve $\frac{2x + 8}{3} = 6$
Step 1: Multiply both sides by 3: $2x + 8 = 18$
Step 2: Subtract 8: $2x = 10$
Step 3: Divide by 2: $x = 5$
Solution: $x = 5$
Step 1: Multiply both sides by 3: $2x + 8 = 18$
Step 2: Subtract 8: $2x = 10$
Step 3: Divide by 2: $x = 5$
Solution: $x = 5$
11. Consecutive Integer Problems
Consecutive Integers: Integers that follow each other in order with a difference of 1
Examples: 5, 6, 7 or -3, -2, -1, 0
Examples: 5, 6, 7 or -3, -2, -1, 0
Representing Consecutive Integers:
Consecutive Integers:
First integer: $n$
Second integer: $n + 1$
Third integer: $n + 2$
Consecutive Even Integers:
First: $n$ (where $n$ is even)
Second: $n + 2$
Third: $n + 4$
Consecutive Odd Integers:
First: $n$ (where $n$ is odd)
Second: $n + 2$
Third: $n + 4$
Consecutive Integers:
First integer: $n$
Second integer: $n + 1$
Third integer: $n + 2$
Consecutive Even Integers:
First: $n$ (where $n$ is even)
Second: $n + 2$
Third: $n + 4$
Consecutive Odd Integers:
First: $n$ (where $n$ is odd)
Second: $n + 2$
Third: $n + 4$
Example 1: "The sum of three consecutive integers is 48. Find the integers."
Let: $n$ = first integer, $n+1$ = second, $n+2$ = third
Equation: $n + (n+1) + (n+2) = 48$
Solve:
$3n + 3 = 48$
$3n = 45$
$n = 15$
Answer: The integers are 15, 16, and 17
Check: $15 + 16 + 17 = 48$ ✓
Let: $n$ = first integer, $n+1$ = second, $n+2$ = third
Equation: $n + (n+1) + (n+2) = 48$
Solve:
$3n + 3 = 48$
$3n = 45$
$n = 15$
Answer: The integers are 15, 16, and 17
Check: $15 + 16 + 17 = 48$ ✓
Example 2: "The sum of two consecutive even integers is 94. Find the integers."
Let: $n$ = first even integer, $n+2$ = second even integer
Equation: $n + (n+2) = 94$
Solve:
$2n + 2 = 94$
$2n = 92$
$n = 46$
Answer: The integers are 46 and 48
Check: $46 + 48 = 94$ ✓
Let: $n$ = first even integer, $n+2$ = second even integer
Equation: $n + (n+2) = 94$
Solve:
$2n + 2 = 94$
$2n = 92$
$n = 46$
Answer: The integers are 46 and 48
Check: $46 + 48 = 94$ ✓
Example 3: "Find three consecutive odd integers whose sum is 75."
Let: $n, n+2, n+4$ be the three consecutive odd integers
Equation: $n + (n+2) + (n+4) = 75$
Solve:
$3n + 6 = 75$
$3n = 69$
$n = 23$
Answer: The integers are 23, 25, and 27
Check: $23 + 25 + 27 = 75$ ✓
Let: $n, n+2, n+4$ be the three consecutive odd integers
Equation: $n + (n+2) + (n+4) = 75$
Solve:
$3n + 6 = 75$
$3n = 69$
$n = 23$
Answer: The integers are 23, 25, and 27
Check: $23 + 25 + 27 = 75$ ✓
12. Solve Linear Equations with Variables on Both Sides
Variables on Both Sides: When the variable appears on both the left and right sides of the equation
Example: $3x + 5 = 2x + 9$
Example: $3x + 5 = 2x + 9$
Steps to Solve:
Step 1: Simplify both sides (distribute, combine like terms)
Step 2: Move all variable terms to one side (usually left)
Step 3: Move all constant terms to the other side
Step 4: Solve for the variable
Step 5: Check your solution
Step 1: Simplify both sides (distribute, combine like terms)
Step 2: Move all variable terms to one side (usually left)
Step 3: Move all constant terms to the other side
Step 4: Solve for the variable
Step 5: Check your solution
Example 1: Solve $5x + 3 = 2x + 15$
Step 1: Subtract $2x$ from both sides
$5x - 2x + 3 = 2x - 2x + 15$
$3x + 3 = 15$
Step 2: Subtract 3 from both sides
$3x = 12$
Step 3: Divide by 3
$x = 4$
Solution: $x = 4$
Check: $5(4) + 3 = 23$ and $2(4) + 15 = 23$ ✓
Step 1: Subtract $2x$ from both sides
$5x - 2x + 3 = 2x - 2x + 15$
$3x + 3 = 15$
Step 2: Subtract 3 from both sides
$3x = 12$
Step 3: Divide by 3
$x = 4$
Solution: $x = 4$
Check: $5(4) + 3 = 23$ and $2(4) + 15 = 23$ ✓
Example 2: Solve $7x - 10 = 3x + 6$
Step 1: Subtract $3x$ from both sides
$4x - 10 = 6$
Step 2: Add 10 to both sides
$4x = 16$
Step 3: Divide by 4
$x = 4$
Solution: $x = 4$
Step 1: Subtract $3x$ from both sides
$4x - 10 = 6$
Step 2: Add 10 to both sides
$4x = 16$
Step 3: Divide by 4
$x = 4$
Solution: $x = 4$
Example 3: Solve $2(3x - 4) = 4(x + 1)$
Step 1: Distribute both sides
$6x - 8 = 4x + 4$
Step 2: Subtract $4x$ from both sides
$2x - 8 = 4$
Step 3: Add 8 to both sides
$2x = 12$
Step 4: Divide by 2
$x = 6$
Solution: $x = 6$
Step 1: Distribute both sides
$6x - 8 = 4x + 4$
Step 2: Subtract $4x$ from both sides
$2x - 8 = 4$
Step 3: Add 8 to both sides
$2x = 12$
Step 4: Divide by 2
$x = 6$
Solution: $x = 6$
Tip: Move the smaller variable term to eliminate it and keep the coefficient positive when possible
13. Solve Linear Equations: Mixed Review
General Strategy for Any Linear Equation:
1. Simplify both sides (distribute, combine like terms)
2. Move variables to one side
3. Move constants to the other side
4. Isolate the variable
5. Check your solution
1. Simplify both sides (distribute, combine like terms)
2. Move variables to one side
3. Move constants to the other side
4. Isolate the variable
5. Check your solution
Practice Problems - Mixed Types:
1. One-Step: $x + 12 = 20$
Solution: $x = 8$
2. Two-Step: $4x - 7 = 21$
Solution: $x = 7$
3. Multi-Step: $3(x - 2) + 5 = 14$
$3x - 6 + 5 = 14$ → $3x - 1 = 14$ → $3x = 15$ → $x = 5$
4. Variables on Both Sides: $5x + 8 = 3x + 20$
$2x + 8 = 20$ → $2x = 12$ → $x = 6$
5. With Fractions: $\frac{x}{2} + 3 = 7$
$\frac{x}{2} = 4$ → $x = 8$
6. Complex: $2(3x + 1) - 5 = 4x + 7$
$6x + 2 - 5 = 4x + 7$ → $6x - 3 = 4x + 7$ → $2x = 10$ → $x = 5$
1. One-Step: $x + 12 = 20$
Solution: $x = 8$
2. Two-Step: $4x - 7 = 21$
Solution: $x = 7$
3. Multi-Step: $3(x - 2) + 5 = 14$
$3x - 6 + 5 = 14$ → $3x - 1 = 14$ → $3x = 15$ → $x = 5$
4. Variables on Both Sides: $5x + 8 = 3x + 20$
$2x + 8 = 20$ → $2x = 12$ → $x = 6$
5. With Fractions: $\frac{x}{2} + 3 = 7$
$\frac{x}{2} = 4$ → $x = 8$
6. Complex: $2(3x + 1) - 5 = 4x + 7$
$6x + 2 - 5 = 4x + 7$ → $6x - 3 = 4x + 7$ → $2x = 10$ → $x = 5$
14. Solve Linear Equations: Complete the Solution
Complete the Solution: Fill in missing steps in a partially solved equation
Example: Complete the solution for $3x + 7 = 22$
$3x + 7 = 22$
$3x + 7 - $ _____ $= 22 - $ _____ (Step 1: Subtract 7)
Answer: 7, 7
$3x = 15$
$\frac{3x}{\_\_\_\_} = \frac{15}{\_\_\_\_}$ (Step 2: Divide by 3)
Answer: 3, 3
$x = $ _____ (Final answer)
Answer: 5
$3x + 7 = 22$
$3x + 7 - $ _____ $= 22 - $ _____ (Step 1: Subtract 7)
Answer: 7, 7
$3x = 15$
$\frac{3x}{\_\_\_\_} = \frac{15}{\_\_\_\_}$ (Step 2: Divide by 3)
Answer: 3, 3
$x = $ _____ (Final answer)
Answer: 5
Example 2: Complete: $2(x - 3) = 10$
$2x - $ _____ $= 10$ (Distribute)
Answer: 6
$2x = $ _____ (Add 6)
Answer: 16
$x = $ _____ (Divide by 2)
Answer: 8
$2x - $ _____ $= 10$ (Distribute)
Answer: 6
$2x = $ _____ (Add 6)
Answer: 16
$x = $ _____ (Divide by 2)
Answer: 8
15. Find the Number of Solutions to a Linear Equation
Types of Solutions:
• One Solution: Equation simplifies to $x = $ (a specific number)
• No Solution: Equation simplifies to a false statement (like $5 = 3$)
• Infinitely Many Solutions: Equation simplifies to a true statement (like $5 = 5$)
• One Solution: Equation simplifies to $x = $ (a specific number)
• No Solution: Equation simplifies to a false statement (like $5 = 3$)
• Infinitely Many Solutions: Equation simplifies to a true statement (like $5 = 5$)
One Solution
When: Variables have different coefficients after simplifying
Result: One unique value for $x$
Result: One unique value for $x$
Example: $3x + 5 = 2x + 8$
$x + 5 = 8$
$x = 3$ ← One solution
$x + 5 = 8$
$x = 3$ ← One solution
No Solution
When: Variables cancel out, leaving a FALSE statement
Result: No value of $x$ works
Result: No value of $x$ works
Example: $2x + 5 = 2x + 8$
Subtract $2x$: $5 = 8$ ← FALSE
No solution (inconsistent equation)
Subtract $2x$: $5 = 8$ ← FALSE
No solution (inconsistent equation)
Infinitely Many Solutions
When: Variables cancel out, leaving a TRUE statement
Result: All real numbers are solutions
Result: All real numbers are solutions
Example: $3x + 6 = 3(x + 2)$
$3x + 6 = 3x + 6$
Subtract $3x$: $6 = 6$ ← TRUE
Infinitely many solutions (identity)
$3x + 6 = 3x + 6$
Subtract $3x$: $6 = 6$ ← TRUE
Infinitely many solutions (identity)
| Type | What Happens | Example Result |
|---|---|---|
| One Solution | Get $x = $ number | $x = 5$ |
| No Solution | Get false statement | $0 = 7$ or $3 = -3$ |
| Infinite Solutions | Get true statement | $0 = 0$ or $5 = 5$ |
16. Create Linear Equations with No Solutions or Infinitely Many Solutions
To Create NO SOLUTION:
Make the variable terms equal but constants different
Form: $ax + b = ax + c$ where $b \neq c$
Make the variable terms equal but constants different
Form: $ax + b = ax + c$ where $b \neq c$
Example: Create an equation with no solution starting with $4x + 7 = $ _____
Answer: $4x + 7 = 4x + 10$ (same variable term, different constant)
Check: $4x - 4x = 10 - 7$ → $0 = 3$ FALSE ✓
Answer: $4x + 7 = 4x + 10$ (same variable term, different constant)
Check: $4x - 4x = 10 - 7$ → $0 = 3$ FALSE ✓
To Create INFINITELY MANY SOLUTIONS:
Make both sides identical after simplifying
Form: $ax + b = ax + b$ (exactly the same)
Make both sides identical after simplifying
Form: $ax + b = ax + b$ (exactly the same)
Example: Create an equation with infinite solutions: $3x - 5 = $ _____
Answer: $3x - 5 = 3(x) - 5$ or $3x - 5 = 3x - 5$
Check: $3x - 3x = -5 + 5$ → $0 = 0$ TRUE ✓
Answer: $3x - 5 = 3(x) - 5$ or $3x - 5 = 3x - 5$
Check: $3x - 3x = -5 + 5$ → $0 = 0$ TRUE ✓
Practice: Fill in the blank:
1. $5x + 2 = 5x + $ _____ to have NO solution
Answer: Any number except 2 (e.g., 7, -3, 0, etc.)
2. $2(3x - 4) = $ _____ to have INFINITE solutions
Answer: $6x - 8$ (expand the left side)
3. $-x + 9 = -x + $ _____ to have INFINITE solutions
Answer: 9
1. $5x + 2 = 5x + $ _____ to have NO solution
Answer: Any number except 2 (e.g., 7, -3, 0, etc.)
2. $2(3x - 4) = $ _____ to have INFINITE solutions
Answer: $6x - 8$ (expand the left side)
3. $-x + 9 = -x + $ _____ to have INFINITE solutions
Answer: 9
17. Solve Linear Equations with Variables on Both Sides: Word Problems
Problem-Solving Strategy:
1. Read and understand the situation
2. Define the variable
3. Write expressions for both scenarios
4. Set up an equation
5. Solve the equation
6. Answer in context
1. Read and understand the situation
2. Define the variable
3. Write expressions for both scenarios
4. Set up an equation
5. Solve the equation
6. Answer in context
Example 1: "Gym A charges $20 per month plus a $50 membership fee. Gym B charges $30 per month with no membership fee. After how many months will the total cost be the same?"
Let: $x$ = number of months
Gym A cost: $20x + 50$
Gym B cost: $30x$
Equation: $20x + 50 = 30x$
Solve:
$50 = 30x - 20x$
$50 = 10x$
$x = 5$
Answer: After 5 months, both gyms cost the same.
Let: $x$ = number of months
Gym A cost: $20x + 50$
Gym B cost: $30x$
Equation: $20x + 50 = 30x$
Solve:
$50 = 30x - 20x$
$50 = 10x$
$x = 5$
Answer: After 5 months, both gyms cost the same.
Example 2: "Sarah has 3 times as many books as John. If Sarah gives John 12 books, they will have the same number. How many books does each person have?"
Let: $x$ = number of books John has
Sarah has: $3x$ books
After exchange:
John: $x + 12$
Sarah: $3x - 12$
Equation: $x + 12 = 3x - 12$
Solve:
$12 + 12 = 3x - x$
$24 = 2x$
$x = 12$
Answer: John has 12 books, Sarah has 36 books.
Let: $x$ = number of books John has
Sarah has: $3x$ books
After exchange:
John: $x + 12$
Sarah: $3x - 12$
Equation: $x + 12 = 3x - 12$
Solve:
$12 + 12 = 3x - x$
$24 = 2x$
$x = 12$
Answer: John has 12 books, Sarah has 36 books.
Example 3: "A cell phone plan costs $40 per month plus $0.10 per text. Another plan costs $25 per month plus $0.25 per text. For how many texts will the plans cost the same?"
Let: $x$ = number of texts
Plan 1: $40 + 0.10x$
Plan 2: $25 + 0.25x$
Equation: $40 + 0.10x = 25 + 0.25x$
Solve:
$40 - 25 = 0.25x - 0.10x$
$15 = 0.15x$
$x = 100$
Answer: At 100 texts, both plans cost the same.
Let: $x$ = number of texts
Plan 1: $40 + 0.10x$
Plan 2: $25 + 0.25x$
Equation: $40 + 0.10x = 25 + 0.25x$
Solve:
$40 - 25 = 0.25x - 0.10x$
$15 = 0.15x$
$x = 100$
Answer: At 100 texts, both plans cost the same.
18. Rearrange Multi-Variable Equations
Multi-Variable Equation: An equation with more than one variable
Solve for a Variable: Isolate one variable in terms of the others (literal equations)
Solve for a Variable: Isolate one variable in terms of the others (literal equations)
Steps to Rearrange (Solve for a Variable):
1. Identify which variable to isolate
2. Use inverse operations to move other terms
3. Treat other variables as constants
4. Isolate the desired variable
1. Identify which variable to isolate
2. Use inverse operations to move other terms
3. Treat other variables as constants
4. Isolate the desired variable
Common Formulas to Rearrange
Example 1: Solve $A = lw$ for $w$ (Area formula)
Divide both sides by $l$:
$\frac{A}{l} = \frac{lw}{l}$
Answer: $w = \frac{A}{l}$
Divide both sides by $l$:
$\frac{A}{l} = \frac{lw}{l}$
Answer: $w = \frac{A}{l}$
Example 2: Solve $P = 2l + 2w$ for $l$ (Perimeter)
Step 1: Subtract $2w$ from both sides
$P - 2w = 2l$
Step 2: Divide by 2
$\frac{P - 2w}{2} = l$
Answer: $l = \frac{P - 2w}{2}$
Step 1: Subtract $2w$ from both sides
$P - 2w = 2l$
Step 2: Divide by 2
$\frac{P - 2w}{2} = l$
Answer: $l = \frac{P - 2w}{2}$
Example 3: Solve $y = mx + b$ for $x$ (Slope-intercept form)
Step 1: Subtract $b$ from both sides
$y - b = mx$
Step 2: Divide by $m$
$\frac{y - b}{m} = x$
Answer: $x = \frac{y - b}{m}$
Step 1: Subtract $b$ from both sides
$y - b = mx$
Step 2: Divide by $m$
$\frac{y - b}{m} = x$
Answer: $x = \frac{y - b}{m}$
Example 4: Solve $ax + by = c$ for $y$
Step 1: Subtract $ax$ from both sides
$by = c - ax$
Step 2: Divide by $b$
$y = \frac{c - ax}{b}$
Answer: $y = \frac{c - ax}{b}$
Step 1: Subtract $ax$ from both sides
$by = c - ax$
Step 2: Divide by $b$
$y = \frac{c - ax}{b}$
Answer: $y = \frac{c - ax}{b}$
Example 5: Solve $C = \frac{5}{9}(F - 32)$ for $F$ (Temperature)
Step 1: Multiply both sides by $\frac{9}{5}$
$\frac{9}{5}C = F - 32$
Step 2: Add 32 to both sides
$\frac{9}{5}C + 32 = F$
Answer: $F = \frac{9}{5}C + 32$
Step 1: Multiply both sides by $\frac{9}{5}$
$\frac{9}{5}C = F - 32$
Step 2: Add 32 to both sides
$\frac{9}{5}C + 32 = F$
Answer: $F = \frac{9}{5}C + 32$
Practice Table
| Original Formula | Solve For | Rearranged Formula |
|---|---|---|
| $d = rt$ | $t$ | $t = \frac{d}{r}$ |
| $I = Prt$ | $r$ | $r = \frac{I}{Pt}$ |
| $A = \frac{1}{2}bh$ | $h$ | $h = \frac{2A}{b}$ |
| $V = lwh$ | $h$ | $h = \frac{V}{lw}$ |
| $3x + 2y = 12$ | $y$ | $y = \frac{12 - 3x}{2}$ |
Quick Reference Guide
Equation-Solving Process:
1. Simplify: Use distributive property, combine like terms
2. Isolate Variable Term: Move variables to one side, constants to other
3. Isolate Variable: Divide/multiply to get variable alone
4. Check: Substitute answer back into original equation
1. Simplify: Use distributive property, combine like terms
2. Isolate Variable Term: Move variables to one side, constants to other
3. Isolate Variable: Divide/multiply to get variable alone
4. Check: Substitute answer back into original equation
Properties of Equality:
• Addition: If $a = b$, then $a + c = b + c$
• Subtraction: If $a = b$, then $a - c = b - c$
• Multiplication: If $a = b$, then $ac = bc$
• Division: If $a = b$, then $\frac{a}{c} = \frac{b}{c}$ ($c \neq 0$)
• Addition: If $a = b$, then $a + c = b + c$
• Subtraction: If $a = b$, then $a - c = b - c$
• Multiplication: If $a = b$, then $ac = bc$
• Division: If $a = b$, then $\frac{a}{c} = \frac{b}{c}$ ($c \neq 0$)
Types of Solutions:
• One Solution: Get $x = $ number
• No Solution: Get false statement (like $5 = 3$)
• Infinite Solutions: Get true statement (like $5 = 5$)
• One Solution: Get $x = $ number
• No Solution: Get false statement (like $5 = 3$)
• Infinite Solutions: Get true statement (like $5 = 5$)
Consecutive Integer Representations:
• Consecutive: $n, n+1, n+2$
• Consecutive Even/Odd: $n, n+2, n+4$
• Consecutive: $n, n+1, n+2$
• Consecutive Even/Odd: $n, n+2, n+4$
Success Tips:
✓ Always perform the same operation on both sides
✓ Work in reverse order of operations (PEMDAS backwards)
✓ Keep equations balanced like a scale
✓ Check your solution by substituting back
✓ Show all steps clearly
✓ Read word problems carefully and define variables
✓ Always perform the same operation on both sides
✓ Work in reverse order of operations (PEMDAS backwards)
✓ Keep equations balanced like a scale
✓ Check your solution by substituting back
✓ Show all steps clearly
✓ Read word problems carefully and define variables