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Chapter 2: Problem 71

Candy. \(\quad\) A candy maker wants to make a 60 -pound mixture of two candies to sell for \(\$ 2\) per pound. If black licorice bits sell for \(\$ 1.90\) per pound and orange gumdrops sell for \(\$ 2.20\) per pound, how many pounds of each should be used?

Short Answer

Expert verified
40 pounds of black licorice bits and 20 pounds of orange gumdrops.

Step by step solution

01

Define Variables

Let \( x \) represent the number of pounds of black licorice bits, and \( y \) the number of pounds of orange gumdrops.
02

Formulate Equations

We have two equations: the total weight equation \( x + y = 60 \) and the total cost equation \( 1.90x + 2.20y = 2 \times 60 = 120 \).
03

Solve System: Substitute

Solve the first equation \( x + y = 60 \) for \( y \): \( y = 60 - x \). Substitute \( y \) into the second equation: \( 1.90x + 2.20(60 - x) = 120 \).
04

Simplify and Solve

Distribute and simplify the equation: \( 1.90x + 132 - 2.20x = 120 \). Simplify to get \( -0.30x + 132 = 120 \).
05

Solve for \( x \)

Subtract 132 from both sides: \( -0.30x = -12 \). Then divide both sides by \(-0.30\): \( x = 40 \).
06

Solve for \( y \)

Use \( y = 60 - x \) to find \( y \): \( y = 60 - 40 = 20 \).
07

Verification

Verify the solution: 40 pounds of black licorice at \\(1.90 each and 20 pounds of orange gumdrops at \\)2.20 each totals \( 1.90 \times 40 + 2.20 \times 20 = 120 \). Both equations are satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Problem Solving with Linear Equations
Problem solving in mathematics often involves understanding what you need to find and how to go about finding it. When dealing with linear equations, as in the candy problem, you start by setting up equations based on the information given. This involves:
  • Defining variables to represent the unknowns.
  • Formulating equations from the problem statement.
  • Systematically solving these equations to find the required values.
In our candy-maker鈥檚 case, the goal is to find out how much of each candy should be mixed to meet a specific weight and price. Both the total weight and total price are critical factors. By setting up equations to represent these factors, you can proceed to solve the problem carefully using systematic methods explained in linear algebra.
Understanding System of Equations in Word Problems
A system of equations involves multiple equations that you solve together to find the values of different variables. In the candy-making problem, you have two equations:
  • The **weight equation**: which helps ensure that the mix weighs exactly 60 pounds.
  • The **cost equation**: which ensures the cost matches the intended price of the mix.
Systems of equations are commonly used in real-world problems because they allow us to capture relationships between different variables. Finding a solution that satisfies both equations simultaneously involves careful substitution or elimination methods. This ensures consistency in meeting both criteria: weight and cost, necessary to solve the problem accurately and efficiently.
The Substitution Method for Solving Systems
The substitution method is a straightforward way to solve systems of equations, especially when one equation can be easily solved for one variable. In our problem, we:
  • Solve the first equation for one variable (e.g., solve for \( y \) in terms of \( x \)).
  • Substitute this expression into the other equation.
  • Simplify the resulting equation to find the value of one variable.
  • Substitute back to find the other variable.
This method is particularly useful when the equations are linear, as it breaks down a potentially complex problem into two simpler steps. By isolating one variable first, you make the process of finding a solution more manageable and less prone to errors. After correctly solving for both variables, always verify your solution by substituting the values back into the original equations to ensure all conditions are met.

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