Problem 46 - Elementary Algebra

Question

A small town library buys a total of 35 books that cost \(\$ 1022\). Some of the books cost \(\$ 22\) each, and the remainder cost \(\$ 34\) per book. How many books of each price did the library buy?

Step-by-Step Solution

Answer

The library bought 14 books at \(\$22\) each and 21 books at \(\$34\) each.

Step 1: Define Variables

Let \( x \) represent the number of books that cost \( \\(22 \) each. Let \( y \) represent the number of books that cost \( \\)34 \) each.

Step 2: Set Up Equation for Total Books

Since the library buys a total of 35 books, we can write the equation: \( x + y = 35 \).

Step 3: Set Up Equation for Total Cost

The cost of the books that cost \( \\(22 \) each is \( 22x \), and the cost of the books that cost \( \\)34 \) each is \( 34y \). The total cost is \( \$1022 \), so the equation is: \( 22x + 34y = 1022 \).

Step 4: Solve the System of Equations

We have the system of equations:1. \( x + y = 35 \)2. \( 22x + 34y = 1022 \)First, solve the first equation for \( y \): \( y = 35 - x \).Substitute \( y = 35 - x \) into the second equation: \( 22x + 34(35 - x) = 1022 \).

Step 5: Simplify and Solve for x

Simplify the equation: \( 22x + 1190 - 34x = 1022 \).Combine like terms: \( -12x + 1190 = 1022 \).Subtract 1190 from both sides: \( -12x = -168 \).Divide both sides by -12: \( x = 14 \).

Step 6: Solve for y

Substitute \( x = 14 \) back into the equation \( y = 35 - x \):\( y = 35 - 14 = 21 \).

Step 7: Verify the Solution

Check that these values satisfy the original conditions:- Total books: \( 14 + 21 = 35 \).- Total cost: \( 22 \times 14 + 34 \times 21 = 308 + 714 = 1022 \).Both conditions are satisfied.

Key Concepts

Algebraic Problem Solving

Understanding algebraic problem solving can significantly simplify complex problems. It's a method of translating real-world situations into mathematical equations, which we can then solve step by step. In this example, the problem is identifying how many books were bought at each price by the library. By breaking down the problem into smaller, more manageable parts, we use algebra to reach a logical solution.

This involves identifying what we know from the problem statement, assigning variables to unknown quantities, and developing equations that tie the known and unknown parts together. Solving these equations, either by substitution or elimination, allows for a step-by-step unraveling of the unknown, bringing clarity to the original problem.

• Translate the problem into mathematical language.
• Break down the equations into smaller steps.
• Solve equations systematically.

Mastering this method enhances critical thinking and problem-solving skills.

Equation Setup

Equation setup is the crucial initial stage where we translate word problems into mathematical expressions. In our problem, the total number of books and total cost provide the foundation for our equations. To properly set this up, begin by establishing straightforward equations based on the relationships established in the problem description.

Two main equations are derived from the problem:

• The total number of books: \( x + y = 35 \).


• The total cost equation: \( 22x + 34y = 1022 \).

These equations are grounded directly in the context given. It's vital to ensure that each equation accurately reflects the quantities and relationships from the problem to solve for unknown variables effectively.

Ultimately, a well-structured equation setup simplifies the problem-solving process and serves as a strong base for further analysis.

Variable Definition

Defining variables is a fundamental step in modeling a problem with mathematical equations. By introducing variables like \(x\) and \(y\), we replace complex, wordy descriptions with cleaner, conceptual placeholders.

In this scenario, we need to find the number of books purchased at two different prices. Thus, we define:

• \( x \) as the number of books costing \( \\(22 \) each.


• \( y \) as the number of books costing \( \\)34 \) each.

Choosing intuitive variable names that clearly represent the quantities they substitute makes these equations easier to understand. This practice is crucial for keeping track of what each variable represents throughout solving the problem. Variable definition sets the stage for establishing clear and solvable equations from complex scenarios, ultimately demystifying real-world algebraic problems.