Question

A lady buys 20 trinkets at a yard sale. The cost of each trinket is either $$\$ 0.30$$ or $$\$ 0.65 .$$ If she spends $$\$ 8.80$$, how many of each type of trinket does she buy?

Short answer

12 trinkets at $0.30 each and 8 trinkets at $0.65 each.

Step-by-step solution

Define the Variables

Let \( x \) be the number of trinkets that cost \\(0.30 each, and \( y \) be the number of trinkets that cost \\)0.65 each.

Set up the Equation for Total Number of Trinkets

Since the lady buys 20 trinkets in total, we have the equation: \[ x + y = 20. \]

Set up the Equation for Total Cost

The total cost spent is \$8.80, so we have the cost equation: \[ 0.30x + 0.65y = 8.80. \]

Solve the Equations Simultaneously

Using the equations from Steps 2 and 3, solve for \( x \) and \( y \).First from \( x + y = 20 \), express \( y \) in terms of \( x \): \[ y = 20 - x. \]Substitute \( y = 20 - x \) into the cost equation: \[ 0.30x + 0.65(20 - x) = 8.80. \]Simplify and solve for \( x \): \[ 0.30x + 13 - 0.65x = 8.80 \]\[ -0.35x + 13 = 8.80 \]\[ -0.35x = 8.80 - 13 \]\[ -0.35x = -4.20 \]\[ x = \frac{-4.20}{-0.35} = 12. \]

Find the Value of y

Substitute \( x = 12 \) back into \( y = 20 - x \) to find \( y \): \[ y = 20 - 12 = 8. \]

Verify the Solution

Double-check the calculations: The lady buys 12 trinkets at \\(0.30 each and 8 trinkets at \\)0.65 each: For the total number of trinkets: \( 12 + 8 = 20 \).For the total cost: \( 12 \times 0.30 + 8 \times 0.65 = 3.60 + 5.20 = 8.80 \), which matches the amount spent.

Key concepts

System of Equations

When you hear the term "system of equations," it refers to a collection of two or more equations with the same set of unknowns. In algebra, these systems can be a powerful tool to solve real-world problems like the one involving trinkets at a yard sale. This approach allows you to find values for each unknown that satisfy all the given conditions.

To solve these equations simultaneously, often you need to:

• Define your variables clearly, like letting \( x \) be the number of cheaper trinkets and \( y \) the pricier ones.
• Write down each condition as a separate equation.
• Use techniques like substitution or elimination to solve these equations together.

These steps lead you to the solution that fits all given conditions in the problem, ensuring you've understood and applied algebra concepts effectively.

Word Problems

Word problems are fascinating because they connect mathematical concepts to real-life scenarios. They often describe situations in which the principles of algebra can be used to find solutions, like determining the number of items someone can purchase given certain budget constraints.

Breaking down word problems typically involves several steps:

• Read the problem thoroughly to understand what is being asked.
• Identify the quantities involved and assign variables to them.
• Translate the words into mathematical statements or equations.
• Solve the resulting equations using appropriate algebraic methods.

By practicing word problems, you can enhance your critical thinking and problem-solving skills, making algebra more approachable and less daunting.

Simultaneous Equations

Simultaneous equations are equations that we need to solve at the same time, meaning we are looking for the same solution set for both. These are common in algebra problems where you need to find multiple unknowns constrained by several conditions, like finding how many trinkets of each type were bought.

The key to solving them often lies in:

• Identifying relations between different variables through given conditions.
• Using one of the equations to express one variable in terms of the other.
• Substituting this expression into the other equation to find the values of both variables.

This simultaneous solving process allows you to handle complex problems easier, ensuring both equations are satisfied by the solutions you reach. It is a versatile tool in algebra that offers clear, methodical solutions to intricate problems.