Problem 36 - Beginning and Intermediate Algebra

Question

Write a system of equations and solve. Mrs. Kowalski bought nine packages of batteries when they were on sale. The AA batteries cost \(\$ 1.00\) per package and the \(C\) batteries cost \(\$ 1.50\) per package. If she spent \(\$ 11.50,\) how many packages of each type of battery did she buy?

Step-by-Step Solution

Answer

Mrs. Kowalski bought 4 packages of AA batteries and 5 packages of C batteries.

Step 1: Define the variables

Let \(x\) represent the number of packages of AA batteries and \(y\) represent the number of packages of C batteries.

Step 2: Write the equations based on the information given

We know Mrs. Kowalski bought a total of 9 packages, so we can write the first equation as \(x+y=9\). We also know the cost of each package: AA batteries cost \(1.00 and C batteries cost \)1.50. She spent a total of $11.50. Therefore, we can write the second equation as \(1x + 1.50y = 11.50\). So, our system of equations is: \(x+y=9\) \(1x + 1.50y = 11.50\)

Step 3: Solve the system of equations

To solve this system of equations, we can use the substitution method. First, we'll solve the first equation for x, then substitute this expression into the second equation. 1. Solve for x in the first equation: \(x=9-y\) 2. Substitute this expression into the second equation: \(1(9-y) + 1.50y = 11.50\) 3. Simplify and solve for y: \(9 - y + 1.50y = 11.50\) \(.50y = 2.50\) \(y = 5\) 4. Now that we have the value of y, we can substitute it back into the equation for x: \(x = 9 - y = 9 - 5 = 4\)

Step 4: Interpret the solution

So, Mrs. Kowalski bought 4 packages of AA batteries (x) and 5 packages of C batteries (y).

Key Concepts

Linear Equations

Linear equations are a fundamental concept in algebra. They are equations of the first degree, meaning they involve variables raised only to the power of one. In simpler terms, they are the mathematical statements that represent a straight line when graphically plotted.

Linear equations generally follow the format: \[ax + by = c\]where:

• \(a\) and \(b\) are constants,
• \(x\) and \(y\) are variables,
• \(c\) is a constant term.

In the exercise, the equations \(x + y = 9\) and \(1x + 1.50y = 11.50\) are both linear. They represent relationships between two variables, specifically the number of battery packages in this scenario. Linear equations like these are essential for modeling real-world situations in algebra.

Algebraic Expressions

Algebraic expressions are a way to represent numbers and relationships using variables and constants connected by mathematical operations like addition, subtraction, and multiplication. They are the building blocks of algebra equations, allowing for dynamic modeling of various scenarios.

Consider the expression \(x + y\), which denotes the total number of battery packages Mrs. Kowalski bought. Here, \(x\) and \(y\) are placeholders for unknown values, making algebraic expressions highly flexible.

Another crucial expression in the solution is \(1x + 1.50y\). It represents the total cost of the packages, with each term accounting for the individual costs of AA and C batteries. By translating word problems into such expressions, complex scenarios are simplified into manageable mathematical elements that are easy to manipulate and solve.

Substitution Method

The substitution method is a technique used to solve a system of equations. This method involves solving one of the equations for one variable and substituting the resulting expression into the other equation. This approach simplifies the equations and makes it easier to find the values of the variables.

In the given problem, we start with \(x + y = 9\) and solve it for \(x\), leading to \(x = 9 - y\). By substituting \(9 - y\) into the second equation \(1x + 1.50y = 11.50\), a single equation in terms of \(y\) is created: \[9 - y + 1.50y = 11.50\]This technique allows focused calculation to find \(y\), which can then be used to find \(x\). The substitution method is particularly useful for systems where the equations are straightforward to manipulate algebraically.

Word Problems in Algebra

Word problems in algebra often describe real-life scenarios that need to be translated into mathematical language to be solved. These problems require critical thinking to identify the relationships and information given in the scenario and convert them into equations.

In the example with Mrs. Kowalski, the key pieces of information are:

• The total number of packages is set at nine.
• The price per package for each type is known.
• The total amount spent is specified.

From this description, one must write algebraic equations that accurately represent these facts. Converting word problems into algebraic expressions and equations allows for a systematic approach to problem-solving in mathematics, turning narrative scenarios into concrete numerical challenges.