Pricing Problem: Pens and Pencils (Set 2 of 5)

Prerequisites

  • Easy - No Calculator
  • Context: Set 2 of 5, a math word problem about prices of pens and pencils.

Problem Statement

  • If I buy two pens and three pencils, it would cost me seven dollars.
  • If I buy five pens, but return a pencil for a full refund, I'll pay a total of nine dollars.
  • Question: What is the dollar value of one pen?
  • Answer choices:
    • A: 0.50
    • B: 1.50
    • C: 2
    • D: 1

Variables

  • Let pp = price of one pen (in dollars).
  • Let cc = price of one pencil (in dollars).

Form the equations

  • From the first statement: 2p+3c=72p + 3c = 7
  • From the second statement: buying five pens and returning one pencil gives a total of nine dollars, so 5p−c=95p - c = 9

Solve the system (Substitution method)

  • From the second equation: 5p−c=9⇒c=5p−95p - c = 9 \Rightarrow c = 5p - 9
  • Substitute into the first equation: 2p+3(5p−9)=72p + 3(5p - 9) = 7
  • Expand and simplify: 2p+15p−27=72p + 15p - 27 = 7
  • Combine like terms: 17p−27=7⇒17p=3417p - 27 = 7 \Rightarrow 17p = 34
  • Solve for pp: p=3417=2p = \frac{34}{17} = 2
  • Compute cc for completeness: c=5p−9=5(2)−9=10−9=1c = 5p - 9 = 5(2) - 9 = 10 - 9 = 1

Solution and answer

  • Price of one pen: p=2p = 2 dollars.
  • Price of one pencil: c=1c = 1 dollar (for completeness).
  • Verification:
    • Check with the first equation: 2p+3c=2(2)+3(1)=4+3=7✓2p + 3c = 2(2) + 3(1) = 4 + 3 = 7\checkmark
    • Check with the second equation: 5p−c=5(2)−1=10−1=9✓5p - c = 5(2) - 1 = 10 - 1 = 9\checkmark
  • Correct option: C (2 dollars).

Key concepts

  • Setting up a system of linear equations to represent real-world price relationships.
  • Unknowns represent item prices; equations model combinations and refunds.
  • Substitution method used to solve for one variable first, then the other.
  • Verification by substituting back into original equations.

Step-by-step reasoning (summary)

  • Translate word problem into two linear equations:
    • 2p+3c=72p + 3c = 7
    • 5p−c=95p - c = 9
  • Solve by isolation: from 5p−c=95p - c = 9, obtain c=5p−9c = 5p - 9.
  • Substitute into 2p+3c=72p + 3c = 7 and solve for pp.
  • After finding p=2p = 2, compute cc to confirm consistency with both equations.

Common interpretations and potential pitfalls

  • The phrase "return a pencil for a full refund" implies the total cost is reduced by the price of one pencil, hence subtracting cc in the second equation (not adding).
  • Double-check algebra when substituting; ensure correct sign during expansion.
  • Always verify solutions by plugging back into all original equations.

Extensions and variations

  • If the numbers were different (e.g., 2p + 3c = 7 and 4p - c = 6), solve similarly and compare results.
  • Solve via elimination method: add multiples of equations to eliminate one variable, then solve for the other.
  • Matrix method: write as (2amp;3 5amp;−1)(p c)=(7 9)\begin{pmatrix}2 & 3\ 5 & -1\end{pmatrix} \begin{pmatrix}p \ c\end{pmatrix} = \begin{pmatrix}7 \ 9\end{pmatrix} and compute using inverse matrix if desired.

Foundational connections

  • Demonstrates how two independent observations about a system determine multiple unknowns.
  • Reflects core principle of linear algebra: intersecting lines in the plane yield a unique solution when lines are not parallel.

Real-world relevance

  • Pricing problems, discounts, refunds, and bundle offers can be modeled with simple linear equations.
  • Validates critical thinking in interpreting word problems into mathematical models.

Quick recap

  • Variables: p→p\to pen price, c→c\to pencil price.
  • Equations: 2p+3c=72p + 3c = 7 and 5p−c=95p - c = 9.
  • Solution: p=2,c=1p = 2,\quad c = 1; verification confirms both equations.
  • Answer: Option C:2.\text{Option } C: 2.