Solved Exercises on Systems of Linear Equations Word Problems in Integrated Math 1

Master systems of equations word problems: modeling real-world situations, setting up equations, and solving through these 5 detailed exercises with comprehensive solutions.

Solution: Exercises 1 to 3
1 Cost Problem
Exercise 1
A store sells pens for $2 each and notebooks for $3 each. A customer buys 8 items for a total of $19. How many pens and notebooks did the customer buy? Solve using a system of equations.
Definition:

Word Problem Modeling: Translating real-world constraints into mathematical equations

Method for word problems:
  1. Read the problem carefully and identify what you need to find
  2. Define variables to represent unknown quantities
  3. Translate the given information into mathematical equations
  4. Solve the system of equations
  5. Verify your solution in the context of the problem
  6. State your answer in complete sentences
Variables
x=pens, y=notebooks
System
x + y = 8, 2x + 3y = 19
Solution
(5, 3)
Step 1: Define variables

Let \(x\) = number of pens bought

Let \(y\) = number of notebooks bought

Step 2: Identify constraints from the problem

Constraint 1: Total number of items is 8 → \(x + y = 8\)

Constraint 2: Total cost is $19 → \(2x + 3y = 19\)

Step 3: Write the system of equations

\[ \begin{cases} x + y = 8 \\ 2x + 3y = 19 \end{cases} \]

Step 4: Solve the system using substitution

From the first equation: \(x = 8 - y\)

Substitute into the second equation: \(2(8 - y) + 3y = 19\)

\(16 - 2y + 3y = 19\)

\(16 + y = 19\)

\(y = 3\)

Step 5: Find x using the first equation

\(x = 8 - y = 8 - 3 = 5\)

Step 6: Verify the solution

Total items: \(x + y = 5 + 3 = 8\) ✓

Total cost: \(2x + 3y = 2(5) + 3(3) = 10 + 9 = 19\) ✓

5 pens, 3 notebooks
Final answer:

The customer bought 5 pens and 3 notebooks.

Applied rules:

• Word Problem Translation: Convert verbal constraints to mathematical equations

• Variable Definition: Clearly define what each variable represents

• Solution Verification: Check that the solution satisfies the original problem constraints

2 Age Problem
Exercise 2
Sarah is twice as old as her brother Tom. The sum of their ages is 30 years. How old are Sarah and Tom? Solve using a system of equations.
Definition:

Age Problems: Word problems involving current or future ages of people

Variables
s=Sarah's age, t=Tom's age
System
s = 2t, s + t = 30
Solution
(20, 10)
Step 1: Define variables

Let \(s\) = Sarah's current age

Let \(t\) = Tom's current age

Step 2: Identify relationships from the problem

Relationship 1: Sarah is twice as old as Tom → \(s = 2t\)

Relationship 2: Sum of their ages is 30 → \(s + t = 30\)

Step 3: Write the system of equations

\[ \begin{cases} s = 2t \\ s + t = 30 \end{cases} \]

Step 4: Solve using substitution

Substitute the first equation into the second: \(2t + t = 30\)

\(3t = 30\)

\(t = 10\)

Step 5: Find Sarah's age

Using the first equation: \(s = 2t = 2(10) = 20\)

Step 6: Verify the solution

Is Sarah twice as old as Tom? \(20 = 2(10) = 20\) ✓

Do their ages sum to 30? \(20 + 10 = 30\) ✓

Sarah: 20 years, Tom: 10 years
Final answer:

Sarah is 20 years old and Tom is 10 years old.

Applied rules:

• Age Relationships: Translate "twice as old" into multiplication

• Substitution Method: Use the equation already solved for one variable

• Verification: Check solution against all original conditions

3 Motion Problem
Exercise 3
Two cars start from the same point. Car A travels north at 60 mph while Car B travels south at 40 mph. When will the cars be 300 miles apart? Solve using a system of equations.
Definition:

Motion Problems: Problems involving distance, rate, and time relationships

Variables
t=time, d_A, d_B distances
System
d_A = 60t, d_B = 40t, d_A + d_B = 300
Solution
t = 3 hours
Step 1: Define variables

Let \(t\) = time in hours after departure

Let \(d_A\) = distance traveled by Car A

Let \(d_B\) = distance traveled by Car B

Step 2: Use the distance formula (distance = rate × time)

Car A: \(d_A = 60t\)

Car B: \(d_B = 40t\)

Step 3: Set up the constraint equation

Since the cars travel in opposite directions, their combined distances equal the separation distance

\(d_A + d_B = 300\)

Step 4: Write the system of equations

\[ \begin{cases} d_A = 60t \\ d_B = 40t \\ d_A + d_B = 300 \end{cases} \]

Step 5: Substitute and solve

Substitute the first two equations into the third:

\(60t + 40t = 300\)

\(100t = 300\)

\(t = 3\)

Step 6: Find distances traveled

Car A: \(d_A = 60(3) = 180\) miles

Car B: \(d_B = 40(3) = 120\) miles

Step 7: Verify the solution

Total distance: \(180 + 120 = 300\) miles ✓

Time: 3 hours

Cars will be 300 miles apart after 3 hours
Final answer:

The cars will be 300 miles apart after 3 hours.

Applied rules:

• Distance Formula: \(d = rt\) (distance = rate × time)

• Motion Problems: When objects move in opposite directions, add distances

• System Reduction: Use substitution to reduce to a single variable

Word Problem Modeling Rules and Methods
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}\)
System of Linear Equations
Define Variables
x=what we need to find
Identify unknowns
Translate Words
"sum" → addition
Convert verbal to algebraic
Solve System
Substitution or elimination
Find the solution
Key definitions:

Word Problem: A real-world situation expressed in words that can be modeled with mathematical equations

Variable: A symbol representing an unknown quantity in the problem

Constraint: A condition or limitation described in the problem that must be satisfied

Modeling: The process of translating a word problem into mathematical equations

Complete methodology:
  1. Read Carefully: Understand what the problem is asking
  2. Identify Unknowns: Determine what quantities need to be found
  3. Define Variables: Assign letters to represent unknown quantities
  4. Find Relationships: Identify how the quantities are related
  5. Set Up Equations: Translate relationships into mathematical equations
  6. Solve System: Use substitution or elimination to find the solution
  7. Check Solution: Verify that the solution makes sense in the context of the problem
  8. State Answer: Express the answer in the context of the original problem
Tip 1: Look for keywords that indicate mathematical operations: "sum," "total," "more than," "less than," "twice," "per."
Tip 2: Organize information by listing known and unknown quantities separately.
Tip 3: Make sure your answer is reasonable in the context of the problem.
Tip 4: Always verify that your solution satisfies all constraints in the original problem.
Common errors: Misidentifying variables, translating words incorrectly, arithmetic mistakes, not checking the solution in context.
Exam preparation: Practice different problem types, memorize key translation words, work on verification skills.
Formulas to know by heart:

• Distance formula: \(d = rt\) (distance = rate × time)

• Total value: \(\text{Number of items} \times \text{Price per item}\)

• Age relationships: "is twice as old" → \(x = 2y\), "sum is" → \(x + y = \text{value}\)

• Mixture problems: \(\text{Amount} \times \text{Concentration} = \text{Pure substance}\)

Solution: Exercises 4 to 5
4 Investment Problem
Exercise 4
A person invests $10,000 in two accounts. One account pays 5% annual interest and the other pays 7% annual interest. At the end of the year, the total interest earned is $600. How much was invested in each account? Solve using a system of equations.
Definition:

Investment Problems: Problems involving money invested at different interest rates

Variables
x=5% account, y=7% account
System
x + y = 10000, 0.05x + 0.07y = 600
Solution
(5000, 5000)
Step 1: Define variables

Let \(x\) = amount invested at 5% interest

Let \(y\) = amount invested at 7% interest

Step 2: Identify constraints

Constraint 1: Total investment is $10,000 → \(x + y = 10000\)

Constraint 2: Total interest is $600 → \(0.05x + 0.07y = 600\)

Step 3: Write the system of equations

\[ \begin{cases} x + y = 10000 \\ 0.05x + 0.07y = 600 \end{cases} \]

Step 4: Solve using substitution

From the first equation: \(x = 10000 - y\)

Substitute into the second equation:

\(0.05(10000 - y) + 0.07y = 600\)

\(500 - 0.05y + 0.07y = 600\)

\(500 + 0.02y = 600\)

\(0.02y = 100\)

\(y = 5000\)

Step 5: Find x

\(x = 10000 - y = 10000 - 5000 = 5000\)

Step 6: Verify the solution

Total investment: \(x + y = 5000 + 5000 = 10000\) ✓

Total interest: \(0.05(5000) + 0.07(5000) = 250 + 350 = 600\) ✓

$5000 at 5%, $5000 at 7%
Final answer:

$5,000 was invested at 5% interest and $5,000 was invested at 7% interest.

Applied rules:

• Interest Formula: Interest = Principal × Rate

• Investment Problems: Total investment = sum of individual investments

• Total Interest: Sum of interests from each account

5 Mixture Problem
Exercise 5
A chemist wants to mix a 20% acid solution with a 50% acid solution to obtain 30 liters of a 30% acid solution. How many liters of each solution should be used? Solve using a system of equations.
Definition:

Mixture Problems: Problems involving combining different concentrations to achieve a desired concentration

Variables
x=20% sol, y=50% sol
System
x + y = 30, 0.2x + 0.5y = 9
Solution
(20, 10)
Step 1: Define variables

Let \(x\) = liters of 20% acid solution

Let \(y\) = liters of 50% acid solution

Step 2: Identify constraints

Constraint 1: Total volume is 30 liters → \(x + y = 30\)

Constraint 2: Pure acid content: \(0.20x + 0.50y = 0.30(30)\)

So: \(0.2x + 0.5y = 9\)

Step 3: Write the system of equations

\[ \begin{cases} x + y = 30 \\ 0.2x + 0.5y = 9 \end{cases} \]

Step 4: Solve using substitution

From the first equation: \(x = 30 - y\)

Substitute into the second equation:

\(0.2(30 - y) + 0.5y = 9\)

\(6 - 0.2y + 0.5y = 9\)

\(6 + 0.3y = 9\)

\(0.3y = 3\)

\(y = 10\)

Step 5: Find x

\(x = 30 - y = 30 - 10 = 20\)

Step 6: Verify the solution

Total volume: \(x + y = 20 + 10 = 30\) liters ✓

Pure acid: \(0.2(20) + 0.5(10) = 4 + 5 = 9\) liters

Percentage: \(\frac{9}{30} = 0.3 = 30\%\) ✓

20 liters of 20% solution, 10 liters of 50% solution
Final answer:

The chemist should use 20 liters of the 20% acid solution and 10 liters of the 50% acid solution.

Applied rules:

• Mixture Formula: Amount × Concentration = Pure Substance

• Volume Constraint: Sum of individual volumes equals total volume

• Concentration Equation: Sum of pure substances equals total pure substance

Systems Word Problems Fundamentals
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}\)
System of Linear Equations
Key definitions:

Word Problem: A mathematical problem expressed in everyday language that requires mathematical modeling to solve

Variable: A symbol (usually a letter) that represents an unknown quantity in a mathematical expression

Constraint: A condition or limitation that must be satisfied by the solution

Modeling: The process of creating mathematical equations to represent a real-world situation

Complete methodology:
  1. Comprehend the Problem: Read and understand what is being asked
  2. Identify Unknowns: Determine what quantities are unknown
  3. Define Variables: Assign variables to represent the unknowns
  4. Translate Information: Convert verbal statements to mathematical equations
  5. Formulate System: Write the system of equations based on constraints
  6. Solve System: Use appropriate method to find the solution
  7. Verify Solution: Check that the solution makes sense in the context
  8. Express Answer: State the answer in the context of the original problem
Tip 1: Look for keywords that indicate mathematical relationships: "sum," "difference," "product," "quotient," "more than," "less than."
Tip 2: Organize information in a table or chart to visualize relationships between quantities.
Tip 3: Always check that your answer is reasonable in the context of the problem.
Tip 4: Make sure your variables are clearly defined and your equations accurately represent the problem.
Applications: Business (profit/cost analysis), physics (motion problems), chemistry (mixture problems), finance (investment problems), engineering (optimization).
Properties: Real-world problems often have constraints that limit the solution set; solutions must be meaningful in the given context.
Essential formulas:

• Distance: \(d = rt\) (distance = rate × time)

• Interest: \(I = Prt\) (interest = principal × rate × time)

• Mixture: \(\text{Amount} \times \text{Concentration} = \text{Pure Substance}\)

• General: \(\text{Total} = \sum(\text{Individual Components})\)

Systems Word Problems Visualization
Exercise 6: Problem Types Comparison
Compare different word problem types:
Cost: 2x + 3y = 19 (pens and notebooks)
Age: s = 2t (Sarah and Tom)
Motion: d_A + d_B = 300 (cars moving apart)

Analysis: The chart shows how different word problem types can be modeled with linear equations.

  • Cost problems: Focus on total quantities and values
  • Age problems: Focus on relationships between ages
  • Motion problems: Focus on distance and time relationships

Questions & Answers

Question: How do I know which variable to assign to which quantity in a word problem?

Answer: When assigning variables, follow these guidelines:

  • Identify what you need to find: The problem usually asks for specific quantities
  • Use meaningful letters: Choose letters that remind you of what they represent (e.g., p for pens, t for time)
  • Keep track of units: Make sure your variables represent the correct type of quantity
  • Be consistent: Once you assign a variable, use it consistently throughout the problem

For example, in a problem about pens and notebooks, you might use:

  • p = number of pens
  • n = number of notebooks

Or you could use x and y, but it's helpful to note what each represents. The key is to be clear about what each variable represents and stay consistent.

Question: What if I set up the equations differently? Will I still get the same answer?

Answer: As long as you correctly translate the relationships in the problem, you will get the same answer regardless of how you set up the equations:

  • Same solution: The mathematical relationships are preserved
  • Different arrangement: You might solve for variables in a different order
  • Equivalent systems: Different arrangements of the same information

For example, if you have a problem about pens and notebooks totaling 8 items, you could write:

  • Equation 1: \(x + y = 8\) (pens + notebooks = 8)
  • Equation 2: \(y = 8 - x\) (equivalent to the first)

Both approaches lead to the same solution. The key is ensuring your equations accurately represent the relationships in the problem.

Question: How do I handle word problems with more than two unknowns?

Answer: For problems with more than two unknowns, you need the same number of equations as unknowns:

  • Three unknowns: Need three equations (3×3 system)
  • Four unknowns: Need four equations (4×4 system)
  • Pattern continues: n unknowns → n equations

For example, if a problem mentions three types of items (pens, pencils, erasers), you might need three equations:

  • Total number of items
  • Total cost
  • Relationship between two of the items

For larger systems, elimination method is generally more efficient than substitution. However, most Integrated Math 1 problems will have two unknowns.

Question: How can I check if my answer is reasonable?

Answer: Use these strategies to check reasonableness:

  • Unit consistency: Do your answers have the correct units?
  • Real-world sense: Does the answer make logical sense?
  • Sign check: Are the values positive when they should be?
  • Size check: Are the values reasonable for the context?

For example:

  • If finding the number of people, the answer should be a positive whole number
  • If finding an age, it should be positive and realistic
  • If finding a price, it should be positive

Always substitute your answers back into the original problem to verify they satisfy all conditions.