Learn on PengiPengi Math (Grade 8)Chapter 4: Systems of Linear Equations

Lesson 5: Real-World Applications of Linear Systems

In this Grade 8 Pengi Math lesson from Chapter 4, students learn to translate real-world scenarios — including break-even points, mixture problems, and perimeter/area situations — into systems of linear equations by defining variables and building algebraic models. Students then select the most effective solving method (Graphing, Substitution, or Elimination) based on the structure of their equations. Finally, they practice interpreting the resulting ordered pair solution in context, connecting the math back to meaningful units like cost, time, or quantity.

Section 1

Applying Elimination to Word Problems

Property

When a real-world problem describes two different combinations of the exact same items resulting in different totals, you can model it using two equations in standard form (Ax+By=CAx + By = C). You can then strategically multiply one or both equations to eliminate a variable and find the unit value of each item.

Examples

  • Word Problem Setup: A store sells 3 shirts and 2 pairs of pants for $85\$85. The next day, it sells 4 shirts and 3 pairs of pants for $120\$120. Find the cost of a single shirt (ss) and a single pair of pants (pp).
  • Writing the Equations:

Equation 1: 3s+2p=853s + 2p = 85
Equation 2: 4s+3p=1204s + 3p = 120

  • Solving with Multiplication: Eliminate pp by finding the LCM of 2 and 3 (which is 6).

Multiply the top equation by 33: 3(3s+2p=85)9s+6p=2553(3s + 2p = 85) \rightarrow 9s + 6p = 255
Multiply the bottom equation by 2-2: 2(4s+3p=120)8s6p=240-2(4s + 3p = 120) \rightarrow -8s - 6p = -240
Add the equations: (9s8s)+(6p6p)=2552401s=15(9s - 8s) + (6p - 6p) = 255 - 240 \rightarrow 1s = 15.
A shirt costs $15\$15.

  • Back-Substitution: Plug s=15s = 15 into the first equation:

3(15)+2p=8545+2p=852p=40p=203(15) + 2p = 85 \rightarrow 45 + 2p = 85 \rightarrow 2p = 40 \rightarrow p = 20.
A pair of pants costs $20\$20.

Explanation

Many real-world transactions—like buying varying quantities of two items at a grocery store—give you total costs without telling you the individual prices. By setting up a system of equations, you can act like a mathematical detective. Scaling up the equations allows you to artificially force the quantity of one item to match (like forcing both scenarios to pretend they bought 6 pairs of pants). Once they match, you subtract the scenarios to completely remove that item from the equation, revealing the exact price of the other!

Section 2

Application: Solving Real-World Systems by Graphing

Property

To solve real-world problems using systems of equations by graphing:

  1. Identify the two unknown quantities and assign variables.
  2. Write two equations based on the given relationships.
  3. Graph both equations on the same coordinate plane.
  4. Find the intersection point and interpret the coordinates in context.

Examples

  • Number Puzzle: Two numbers have a sum of 12 and a difference of 2. Graph the system x+y=12x + y = 12 and xy=2x - y = 2. The intersection point (7,5)(7, 5) means the numbers are 7 and 5.
  • Coffee Shop: Premium beans cost 12 per pound and regular beans cost 8 per pound. A customer buys 10 pounds total for 92. Graph the system x+y=10x + y = 10 and 12x+8y=9212x + 8y = 92. The intersection point (3,7)(3, 7) means 3 pounds of premium and 7 pounds of regular beans were bought.
  • Geometry: A rectangle has a perimeter of 20 inches, and the length is 4 inches more than the width. Graph the system 2L+2W=202L + 2W = 20 and L=W+4L = W + 4. The intersection point (7,3)(7, 3) means the length is 7 inches and the width is 3 inches.

Explanation

Real-world problems often involve two unknown quantities with two entirely different relationships between them (like counting total items vs. counting total money). Each relationship becomes its own equation. The intersection point of the graphed lines gives the unique values that satisfy both conditions simultaneously.

Section 3

Solving Real-World Systems by Substitution

Property

Many real-world situations can be modeled by a system of linear equations. The process involves:

  1. Defining variables to represent the unknown quantities.
  2. Writing two linear equations that describe the relationships between the variables.
  3. Solving the system using the substitution method to find the solution.

Examples

Book overview

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Chapter 4: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Introduction to Systems and Solving by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems by Substitution

  3. Lesson 3

    Lesson 3: Solving Systems by Elimination

  4. Lesson 4

    Lesson 4: Number of Solutions and Special Cases

  5. Lesson 5Current

    Lesson 5: Real-World Applications of Linear Systems

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

Applying Elimination to Word Problems

Property

When a real-world problem describes two different combinations of the exact same items resulting in different totals, you can model it using two equations in standard form (Ax+By=CAx + By = C). You can then strategically multiply one or both equations to eliminate a variable and find the unit value of each item.

Examples

  • Word Problem Setup: A store sells 3 shirts and 2 pairs of pants for $85\$85. The next day, it sells 4 shirts and 3 pairs of pants for $120\$120. Find the cost of a single shirt (ss) and a single pair of pants (pp).
  • Writing the Equations:

Equation 1: 3s+2p=853s + 2p = 85
Equation 2: 4s+3p=1204s + 3p = 120

  • Solving with Multiplication: Eliminate pp by finding the LCM of 2 and 3 (which is 6).

Multiply the top equation by 33: 3(3s+2p=85)9s+6p=2553(3s + 2p = 85) \rightarrow 9s + 6p = 255
Multiply the bottom equation by 2-2: 2(4s+3p=120)8s6p=240-2(4s + 3p = 120) \rightarrow -8s - 6p = -240
Add the equations: (9s8s)+(6p6p)=2552401s=15(9s - 8s) + (6p - 6p) = 255 - 240 \rightarrow 1s = 15.
A shirt costs $15\$15.

  • Back-Substitution: Plug s=15s = 15 into the first equation:

3(15)+2p=8545+2p=852p=40p=203(15) + 2p = 85 \rightarrow 45 + 2p = 85 \rightarrow 2p = 40 \rightarrow p = 20.
A pair of pants costs $20\$20.

Explanation

Many real-world transactions—like buying varying quantities of two items at a grocery store—give you total costs without telling you the individual prices. By setting up a system of equations, you can act like a mathematical detective. Scaling up the equations allows you to artificially force the quantity of one item to match (like forcing both scenarios to pretend they bought 6 pairs of pants). Once they match, you subtract the scenarios to completely remove that item from the equation, revealing the exact price of the other!

Section 2

Application: Solving Real-World Systems by Graphing

Property

To solve real-world problems using systems of equations by graphing:

  1. Identify the two unknown quantities and assign variables.
  2. Write two equations based on the given relationships.
  3. Graph both equations on the same coordinate plane.
  4. Find the intersection point and interpret the coordinates in context.

Examples

  • Number Puzzle: Two numbers have a sum of 12 and a difference of 2. Graph the system x+y=12x + y = 12 and xy=2x - y = 2. The intersection point (7,5)(7, 5) means the numbers are 7 and 5.
  • Coffee Shop: Premium beans cost 12 per pound and regular beans cost 8 per pound. A customer buys 10 pounds total for 92. Graph the system x+y=10x + y = 10 and 12x+8y=9212x + 8y = 92. The intersection point (3,7)(3, 7) means 3 pounds of premium and 7 pounds of regular beans were bought.
  • Geometry: A rectangle has a perimeter of 20 inches, and the length is 4 inches more than the width. Graph the system 2L+2W=202L + 2W = 20 and L=W+4L = W + 4. The intersection point (7,3)(7, 3) means the length is 7 inches and the width is 3 inches.

Explanation

Real-world problems often involve two unknown quantities with two entirely different relationships between them (like counting total items vs. counting total money). Each relationship becomes its own equation. The intersection point of the graphed lines gives the unique values that satisfy both conditions simultaneously.

Section 3

Solving Real-World Systems by Substitution

Property

Many real-world situations can be modeled by a system of linear equations. The process involves:

  1. Defining variables to represent the unknown quantities.
  2. Writing two linear equations that describe the relationships between the variables.
  3. Solving the system using the substitution method to find the solution.

Examples

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Introduction to Systems and Solving by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems by Substitution

  3. Lesson 3

    Lesson 3: Solving Systems by Elimination

  4. Lesson 4

    Lesson 4: Number of Solutions and Special Cases

  5. Lesson 5Current

    Lesson 5: Real-World Applications of Linear Systems