39. Linear equations applied to arithmetic contexts

39. Linear equations applied to arithmetic contexts (e.g., age problems) problems - step by step solution procedure and detailed explanation for each problem.

Below are 10 linear equation word problems applied to arithmetic contexts, focusing on scenarios like age problems, number problems, money, and work rates. Each problem includes a step-by-step solution procedure and a detailed explanation to illustrate how linear equations model these situations. The problems are designed to vary in complexity and context while remaining accessible, with clear setups and solutions using one or two variables. All solutions are verified to ensure correctness.

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Problem 1: Age Problem

Five years ago, Alice was twice as old as Bob. Now, Alice is 25 years old. How old is Bob now?

Step-by-Step Solution:

1. Define variables:
   • Let ( B ) be Bob’s current age.
   • Alice’s current age = 25.
2. Set up the equation:
   • Five years ago, Alice’s age =

     25 - 5 = 20

     .
   • Five years ago, Bob’s age =

     B - 5

     .
   • Alice was twice as old as Bob:

     20 = 2(B - 5)

     .
3. Solve the equation:
   • 20 = 2(B - 5)
   • Divide by 2:

     10 = B - 5
   • Add 5:

     B = 15
4. Verify:
   • Five years ago: Bob’s age =

     15 - 5 = 10

     , Alice’s age = 20.
   • Check:

     20 = 2 \cdot 10

     , which is true.

Final Answer:

Bob is 15 years old.

Detailed Explanation:

• Linear Equation Source: The relationship between ages at a specific time (five years ago) creates a linear equation.
• Variable Choice: Using Bob’s current age simplifies the setup since Alice’s current age is given.
• Common Mistake: Misinterpreting “five years ago” (e.g., not subtracting 5 from both ages).
• Context Check: Ages are positive and realistic, and the solution satisfies the condition.

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Problem 2: Number Problem

The sum of two numbers is 45, and one number is 7 more than the other. Find the numbers.

Step-by-Step Solution:

1. Define variables:
   • Let ( x ) be the smaller number.
   • The larger number =

     x + 7

     .
2. Set up the equation:
   • Sum:

     x + (x + 7) = 45

     .
3. Solve the equation:
   • Simplify:

     2x + 7 = 45
   • Subtract 7:

     2x = 38
   • Divide by 2:

     x = 19
   • Larger number:

     x + 7 = 19 + 7 = 26
4. Verify:
   • Sum:

     19 + 26 = 45

     .
   • Difference:

     26 - 19 = 7

     .

Final Answer:

The numbers are 19 and 26.

Detailed Explanation:

• Linear Equation Source: The sum and the relationship between the numbers form a linear equation.
• Variable Choice: Defining one number and expressing the other in terms of it reduces to one equation.
• Common Mistake: Setting up the difference incorrectly (e.g.,

  x - 7

  ).
• Context Check: Both numbers are positive and satisfy both conditions.

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Problem 3: Money Problem

A child has 12 coins, consisting of nickels (5 cents) and dimes (10 cents), with a total value of 95 cents. How many of each coin does the child have?

Step-by-Step Solution:

1. Define variables:
   • Let ( n ) be the number of nickels.
   • Let ( d ) be the number of dimes.
2. Set up equations:
   • Total coins:

     n + d = 12

     .
   • Total value (in cents):

     5n + 10d = 95

     .
3. Solve the system:
   • From

     n + d = 12

     , express

     n = 12 - d

     .
   • Substitute into value equation:

     5(12 - d) + 10d = 95

     .
   • Simplify:

     60 - 5d + 10d = 95
   • 60 + 5d = 95
   • Subtract 60:

     5d = 35
   • Divide by 5:

     d = 7
   • Find ( n ):

     n = 12 - 7 = 5

     .
4. Verify:
   • Coins:

     5 + 7 = 12

     .
   • Value:

     5 \cdot 5 + 10 \cdot 7 = 25 + 70 = 95

     .

Final Answer:

5 nickels and 7 dimes.

Detailed Explanation:

• Linear Equation Source: The number of coins and their total value create a system of linear equations.
• Substitution Method: Expressing one variable in terms of the other simplifies solving.
• Common Mistake: Using dollars instead of cents or misaligning units.
• Context Check: Non-negative integers for coins and correct total value.

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Problem 4: Work Rate Problem

Anna can paint a room in 6 hours, and Bob can paint it in 8 hours. How long will it take them to paint the room together?

Step-by-Step Solution:

1. Define variables:
   • Let ( t ) be the time (in hours) they take together.
2. Set up the equation:
   • Anna’s rate:

     \frac{1}{6}

     room per hour.
   • Bob’s rate:

     \frac{1}{8}

     room per hour.
   • Combined rate:

     \frac{1}{6} + \frac{1}{8} = \frac{1}{t}

     .
3. Solve the equation:
   • Combine rates:

     \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}

     .
   • Equation:

     \frac{7}{24} = \frac{1}{t}

     .
   • Solve:

     t = \frac{24}{7} \approx 3.43

     .
4. Verify:
   • In

     \frac{24}{7}

     hours:
     • Anna paints:

       \frac{24/7}{6} = \frac{24}{42} = \frac{4}{7}

       .
     • Bob paints:

       \frac{24/7}{8} = \frac{24}{56} = \frac{3}{7}

       .
     • Total:

       \frac{4}{7} + \frac{3}{7} = 1

       room.

Final Answer:

They take

\frac{24}{7}

hours (approximately 3.43 hours).

Detailed Explanation:

• Linear Equation Source: Combined work rates equal the rate of completing one job.
• Rate Addition: Rates are added because they work simultaneously.
• Common Mistake: Averaging times (e.g.,

  \frac{6 + 8}{2} = 7

  ) ignores rate-based work.
• Context Check: The time is positive and less than the faster individual time.

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Problem 5: Age Difference

The sum of the ages of a father and son is 48 years. The father is three times as old as the son. How old are they?

Step-by-Step Solution:

1. Define variables:
   • Let ( s ) be the son’s age.
   • Father’s age = ( 3s ).
2. Set up the equation:
   • Sum:

     s + 3s = 48

     .
3. Solve the equation:
   • Simplify:

     4s = 48
   • Divide by 4:

     s = 12
   • Father’s age:

     3 \cdot 12 = 36

     .
4. Verify:
   • Sum:

     12 + 36 = 48

     .
   • Ratio:

     36 = 3 \cdot 12

     .

Final Answer:

Son is 12 years old, father is 36 years old.

Detailed Explanation:

• Linear Equation Source: The sum of ages and the ratio form a linear equation.
• Variable Choice: Using the son’s age simplifies the ratio expression.
• Common Mistake: Misinterpreting “three times as old” as a difference.
• Context Check: Ages are positive integers and realistic.

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Problem 6: Ticket Sales

Adult tickets cost $8, and child tickets cost $5. A group buys 10 tickets for $65. How many of each type were bought?

Step-by-Step Solution:

1. Define variables:
   • Let ( a ) be the number of adult tickets.
   • Let ( c ) be the number of child tickets.
2. Set up equations:
   • Total tickets:

     a + c = 10

     .
   • Total cost:

     8a + 5c = 65

     .
3. Solve the system:
   • From

     a + c = 10

     ,

     a = 10 - c

     .
   • Substitute:

     8(10 - c) + 5c = 65

     .
   • Simplify:

     80 - 8c + 5c = 65
   • 80 - 3c = 65
   • Subtract 80:

     -3c = -15
   • Divide by

     -3

     :

     c = 5
   • Find ( a ):

     a = 10 - 5 = 5

     .
4. Verify:
   • Tickets:

     5 + 5 = 10

     .
   • Cost:

     8 \cdot 5 + 5 \cdot 5 = 40 + 25 = 65

     .

Final Answer:

5 adult tickets and 5 child tickets.

Detailed Explanation:

• Linear Equation Source: The number and cost of tickets form a system of equations.
• Substitution Method: Eliminates one variable for straightforward solving.
• Common Mistake: Mixing up adult and child ticket costs.
• Context Check: Non-negative integers and correct total cost.

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Problem 7: Consecutive Numbers

The sum of three consecutive integers is 72. Find the integers.

Step-by-Step Solution:

1. Define variables:
   • Let the first integer be ( n ).
   • The others are

     n + 1

     ,

     n + 2

     .
2. Set up the equation:
   • Sum:

     n + (n + 1) + (n + 2) = 72

     .
3. Solve the equation:
   • Simplify:

     3n + 3 = 72
   • Subtract 3:

     3n = 69
   • Divide by 3:

     n = 23
   • Integers: ( 23 ), ( 24 ), ( 25 ).
4. Verify:
   • Sum:

     23 + 24 + 25 = 72

     .

Final Answer:

The integers are 23, 24, and 25.

Detailed Explanation:

• Linear Equation Source: The sum of consecutive integers is linear.
• Variable Choice: The first integer simplifies the sequence.
• Common Mistake: Using non-consecutive increments (e.g.,

  n + 2

  ).
• Context Check: Integers are consecutive and sum correctly.

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Problem 8: Investment Problem

A person invests $10,000 in two accounts: one at 4% annual interest and one at 6%. The total interest after one year is $520. How much was invested in each account?

Step-by-Step Solution:

1. Define variables:
   • Let ( x ) be the amount invested at 4%.
   • Amount at 6% =

     10,000 - x

     .
2. Set up the equation:
   • Interest:

     0.04x + 0.06(10,000 - x) = 520

     .
3. Solve the equation:
   • Simplify:

     0.04x + 600 - 0.06x = 520
   • Combine:

     -0.02x + 600 = 520
   • Subtract 600:

     -0.02x = -80
   • Divide by

     -0.02

     :

     x = 4,000
   • Amount at 6%:

     10,000 - 4,000 = 6,000

     .
4. Verify:
   • Interest:

     0.04 \cdot 4,000 + 0.06 \cdot 6,000 = 160 + 360 = 520

     .

Final Answer:

$4,000 at 4%, $6,000 at 6%.

Detailed Explanation:

• Linear Equation Source: The total interest from two investments is linear.
• Variable Choice: One investment amount simplifies the setup.
• Common Mistake: Using percentages without converting to decimals.
• Context Check: Amounts are positive and total $10,000.

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Problem 9: Distance Problem

Two cars leave the same point traveling in opposite directions. One travels at 60 km/h, the other at 80 km/h. After how long will they be 420 km apart?

Step-by-Step Solution:

1. Define variables:
   • Let ( t ) be the time (in hours).
2. Set up the equation:
   • Relative speed =

     60 + 80 = 140

     km/h.
   • Distance:

     140t = 420

     .
3. Solve the equation:
   • Divide by 140:

     t = \frac{420}{140} = 3

     .
4. Verify:
   • Distance:

     60 \cdot 3 + 80 \cdot 3 = 180 + 240 = 420

     .

Final Answer:

They are 420 km apart after 3 hours.

Detailed Explanation:

• Linear Equation Source: The combined speed and distance form a linear equation.
• Relative Speed: Added because the cars travel in opposite directions.
• Common Mistake: Using individual distances without summing.
• Context Check: Time is positive and realistic.

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Problem 10: Mixture Problem

A store sells two types of coffee: one at $6 per pound and one at $9 per pound. How many pounds of each should be mixed to make 10 pounds of a blend worth $7.20 per pound?

Step-by-Step Solution:

1. Define variables:
   • Let ( x ) be the pounds of $6 coffee.
   • Pounds of $9 coffee =

     10 - x

     .
2. Set up the equation:
   • Total value:

     6x + 9(10 - x) = 7.20 \cdot 10

     .
   • Simplify:

     6x + 90 - 9x = 72

     .
3. Solve the equation:
   • Combine:

     -3x + 90 = 72
   • Subtract 90:

     -3x = -18
   • Divide by

     -3

     :

     x = 6
   • Pounds at $9:

     10 - 6 = 4

     .
4. Verify:
   • Weight:

     6 + 4 = 10

     .
   • Value:

     6 \cdot 6 + 9 \cdot 4 = 36 + 36 = 72

     .
   • Price per pound:

     \frac{72}{10} = 7.20

     .

Final Answer:

6 pounds of $6 coffee, 4 pounds of $9 coffee.

Detailed Explanation:

• Linear Equation Source: The total value of the mixture is linear.
• Variable Choice: One type of coffee simplifies the equation.
• Common Mistake: Setting up the value equation incorrectly.
• Context Check: Weights are non-negative and total 10 pounds.

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Summary of Key Concepts:

• Linear Equations: Model relationships like sums, differences, rates, or values in arithmetic contexts.
• Variable Definition: Choose variables to represent unknowns (e.g., ages, quantities).
• Equation Setup: Translate word problem conditions into equations based on sums, rates, or totals.
• Solution Methods: Use substitution for systems or direct solving for single equations.
• Verification: Always check solutions against all conditions.
• Context Check: Ensure solutions are realistic (e.g., positive ages, integer coins).

These problems cover a range of arithmetic contexts, demonstrating how linear equations solve practical problems.