Solved Exercises on Money Word Problems Algebra 1 in Grade 3

Master basic algebraic thinking in money calculations: variables, equations, and problem-solving through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Variable Problem
Exercise 1
Emma had some money in her piggy bank. She added $3.50 and now has $8.75. How much money did she have at first? Let x = money at first.
Definition:

Variable: A symbol that represents an unknown number

Basic algebraic method:
  1. Identify what you know and what you need to find
  2. Assign a variable to the unknown quantity
  3. Set up an equation showing the relationship
  4. Solve by performing the inverse operation
  5. Check your answer
Initial Amount
x
Added Amount
+$3.50
Final Amount
$8.75

Equation: x + $3.50 = $8.75

Step 1: Write the equation

x + $3.50 = $8.75

Step 2: Subtract $3.50 from both sides

x + $3.50 - $3.50 = $8.75 - $3.50

x = $5.25

Step 3: Check the answer

$5.25 + $3.50 = $8.75 ✓

x = $5.25
Final answer:

Emma had $5.25 at first

Applied rules:

• Balance Rule: Perform the same operation on both sides

• Inverse Operation: Subtraction undoes addition

• Verification: Substitute answer to check

2 Subtraction Equation
Exercise 2
Marcus had some money. He spent $4.25 on a snack and now has $2.75 left. How much money did he start with? Let y = starting money.
Definition:

Subtraction Equation: An equation where the variable is decreased by a known amount

Starting Money
y
Amount Spent
-$4.25
Remaining
$2.75

Equation: y - $4.25 = $2.75

Step 1: Write the equation

y - $4.25 = $2.75

Step 2: Add $4.25 to both sides

y - $4.25 + $4.25 = $2.75 + $4.25

y = $7.00

Step 3: Check the answer

$7.00 - $4.25 = $2.75 ✓

y = $7.00
Final answer:

Marcus started with $7.00

Applied rules:

• Inverse Operations: Addition undoes subtraction

• Balance Rule: Same operation on both sides

• Verification: Substitute to confirm

3 Two-step Problem
Exercise 3
Sarah earned some money from chores. She spent $2.50 and then earned $4.75 more. If she now has $9.00, how much did she earn from chores? Let z = money from chores.
Definition:

Two-step Equation: An equation requiring two operations to solve

Started With
z
Spent
-$2.50
Earned More
+$4.75
Final Amount
$9.00

Equation: z - $2.50 + $4.75 = $9.00

Step 1: Write the equation

z - $2.50 + $4.75 = $9.00

Step 2: Simplify the constants

z + ($4.75 - $2.50) = $9.00

z + $2.25 = $9.00

Step 3: Subtract $2.25 from both sides

z + $2.25 - $2.25 = $9.00 - $2.25

z = $6.75

Step 4: Check the answer

$6.75 - $2.50 + $4.75 = $4.25 + $4.75 = $9.00 ✓

z = $6.75
Final answer:

Sarah earned $6.75 from chores

Applied rules:

• Combine Like Terms: Add or subtract constants first

• Balance Rule: Same operation on both sides

• Verification: Substitute to confirm

Algebra 1 Money Methods
\(\text{Variable} + \text{Known} = \text{Total}\)
Addition Equation
\(\text{Variable} - \text{Known} = \text{Total}\)
Subtraction Equation
Addition
x + a = b
Solve by subtracting a
Subtraction
x - a = b
Solve by adding a
Two-step
x + a - b = c
Simplify first, then solve
variable
Variable = Unknown amount
balance
Both sides equal
Solution: Exercises 4 to 5
4 Multiple Purchases
Exercise 4
Tom had some money. He bought a book for $3.50 and a pen for $1.25. If he has $4.75 left, how much money did he start with? Let m = starting money.
Definition:

Multiple Expenses: An equation with multiple subtractions

Starting Money
m
Book Cost
-$3.50
Pen Cost
-$1.25
Remaining
$4.75

Equation: m - $3.50 - $1.25 = $4.75

Step 1: Write the equation

m - $3.50 - $1.25 = $4.75

Step 2: Combine the expenses

$3.50 + $1.25 = $4.75

So: m - $4.75 = $4.75

Step 3: Add $4.75 to both sides

m - $4.75 + $4.75 = $4.75 + $4.75

m = $9.50

Step 4: Check the answer

$9.50 - $3.50 - $1.25 = $9.50 - $4.75 = $4.75 ✓

m = $9.50
Final answer:

Tom started with $9.50

Applied rules:

• Combine Like Terms: Add all expenses together

• Balance Rule: Same operation on both sides

• Verification: Substitute to confirm

5 Repeated Addition
Exercise 5
Lisa saved the same amount each day for 4 days. Then she spent $3.00 on candy. If she has $13.00 left, how much did she save each day? Let w = daily savings.
Definition:

Repeated Addition: Multiplication represented as repeated addition

Daily Savings
w
Days Saved
4
Total Saved
4w
Amount Spent
-$3.00
Remaining
$13.00

Equation: 4w - $3.00 = $13.00

Step 1: Write the equation

4w - $3.00 = $13.00

Step 2: Add $3.00 to both sides

4w - $3.00 + $3.00 = $13.00 + $3.00

4w = $16.00

Step 3: Divide both sides by 4

(4w) ÷ 4 = $16.00 ÷ 4

w = $4.00

Step 4: Check the answer

4 × $4.00 - $3.00 = $16.00 - $3.00 = $13.00 ✓

w = $4.00 per day
Final answer:

Lisa saved $4.00 each day

Applied rules:

• Repeated Addition: 4 days of w = 4w

• Division: To isolate variable multiplied by number

• Verification: Substitute to confirm

Money Algebra 1 Strategies
\(\text{Variable} + \text{Known} = \text{Total} \Rightarrow \text{Variable} = \text{Total} - \text{Known}\)
Solving Addition Equations
Key definitions:

Variable: A symbol representing an unknown value (often x, y, or z)

Equation: A mathematical statement showing two expressions are equal

Balance Rule: Perform the same operation on both sides of the equation

Isolation: Getting the variable alone on one side of the equation

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

Algebra 1 problem-solving methodology:
  1. Identify the unknown: Decide what the variable represents
  2. Set up the equation: Write a mathematical sentence
  3. Solve systematically: Use inverse operations to isolate the variable
  4. Check your answer: Substitute back into the original problem
Tip 1: Always define what your variable represents in the context of the problem.
Tip 2: Addition equations are solved by subtracting the known amount from both sides.
Tip 3: Subtraction equations are solved by adding the known amount to both sides.
Tip 4: Always verify your answer by substituting it back into the original problem.
Common errors: Forgetting to perform the same operation on both sides, misidentifying the unknown, incorrect arithmetic.
Practice strategies: Start with simple equations, work up to complex ones, always check answers.
Algebra 1 formulas for money problems:

• Addition: x + a = b → x = b - a

• Subtraction: x - a = b → x = b + a

• Multiplication: nx = b → x = b ÷ n

• Balance Rule: Whatever you do to one side, do to the other

Questions & Answers

Question: Why do we need to learn about variables when we could just figure out the answer without using letters?

Answer: Variables help us organize our thinking and solve problems systematically:

  • Variables help us clearly identify what we're trying to find
  • They allow us to write the relationship between known and unknown amounts
  • As problems get more complex, variables help us keep track of all the information
  • They prepare us for more advanced math in the future

For example: "I had some money, spent $3, and now have $5" becomes clearer as: x - $3 = $5, where x = $8.

Question: How can I help my child understand the concept of balancing equations?

Answer: Use the seesaw analogy to explain equation balancing:

  • Imagine a seesaw that must stay level
  • Whatever you do to one side, you must do the same thing to the other side
  • If you add weight to one end, you must add the same weight to the other end
  • This keeps the seesaw balanced, just like an equation stays equal

For example: If x + $2 = $7, to find x, we subtract $2 from both sides: x = $7 - $2 = $5.

Question: What's the best way to introduce variables to students who are new to algebra?

Answer: Start with familiar concepts and gradually introduce variables:

  • Begin with fill-in-the-blank problems: "___ + $3 = $8"
  • Then introduce the idea that a letter can represent the blank: "x + $3 = $8"
  • Emphasize that the variable is just a placeholder for a number we don't know yet
  • Always connect the variable back to the real-world context of the problem

For example: "Sarah had some money (x). She got $5 more. Now she has $12. So x + $5 = $12."

Comprehensive Summary: Money Word Problems Algebra 1
Key Concepts and Definitions:

Variable: A symbol (usually a letter) that represents an unknown number (e.g., x, y, z)

Equation: A mathematical statement showing two expressions are equal

Balance Rule: Whatever operation you perform on one side of the equation, you must perform on the other side

Isolation: The process of getting the variable alone on one side of the equation

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

Core Rules and Principles:

• Balance Rule: Maintain equality by performing identical operations on both sides

• Inverse Operations: Use opposite operations to isolate variables

• Verification: Always substitute your answer back to check accuracy

• Define Variables: Clearly state what each variable represents

Step-by-Step Algebra 1 Method:
  1. Identify Unknown: Decide what the variable represents
  2. Set Up Equation: Write the mathematical relationship
  3. Solve Systematically: Use inverse operations to isolate the variable
  4. Check Answer: Substitute back into original equation
Tip 1: Always define what your variable represents in the context of the problem.
Tip 2: Addition equations are solved by subtracting the known amount from both sides.
Tip 3: Subtraction equations are solved by adding the known amount to both sides.
Tip 4: Always verify your answer by substituting it back into the original problem.
Common Pitfalls: Forgetting to perform operations on both sides, misidentifying the unknown, incorrect arithmetic, skipping verification.
Memorization Aids: Remember "what you do to one side, do to the other" and "opposite operations cancel."
Key Takeaways for Students:

• Variables represent unknown amounts in mathematical equations

• Keep equations balanced by doing the same thing to both sides

• Use opposite operations to isolate variables (addition undoes subtraction)

• Always check your answer by substituting it back into the original problem

• Define what each variable represents in the context of the problem