Problem 205 - Elementary Algebra

Question

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for \(\$ 116\). Lisa bought two notebooks and three thumb dives for \(\$ 68\). Find the cost of each notebook and each thumb drive.

Step-by-Step Solution

Answer

Each notebook costs \( \)4 and each thumb drive costs \( \)20.

Step 1: Define Variables

Let the cost of each notebook be denoted as \( n \) dollars and the cost of each thumb drive be denoted as \( t \) dollars.

Step 2: Set Up Equations

From the given information, write the equations based on the purchases. \[ 4n + 5t = 116 \] (Equation 1 - Troy's purchases) \[ 2n + 3t = 68 \] (Equation 2 - Lisa's purchases)

Step 3: Solve the System of Equations

Solve the system using the method of substitution or elimination. We'll use elimination here. First, multiply Equation 2 by 2 to align the coefficients of \( n \). \[ 4n + 6t = 136 \] (Equation 3)

Step 4: Eliminate One Variable

Subtract Equation 1 from Equation 3 to eliminate \( n \): \[ (4n + 6t) - (4n + 5t) = 136 - 116 \] \[ t = 20 \]

Step 5: Solve for the Remaining Variable

With \( t \) known, substitute \( t = 20 \) back into one of the original equations (Equation 2): \[ 2n + 3(20) = 68 \] \[ 2n + 60 = 68 \] \[ 2n = 8 \] \[ n = 4 \]

Step 6: Verify the Solution

Substitute \( n = 4 \) and \( t = 20 \) into both original equations to ensure correctness. \[ 4(4) + 5(20) = 16 + 100 = 116 \] (Equation 1 is satisfied) \[ 2(4) + 3(20) = 8 + 60 = 68 \] (Equation 2 is satisfied)

Key Concepts

defining variables

When solving a system of equations, the first step is to clearly define the variables. In this problem, we need to determine the cost of each notebook and each thumb drive. We make this easier by using letters to represent unknown values. Let’s denote the cost of each notebook as \( n \) dollars and the cost of each thumb drive as \( t \) dollars. By defining variables, we can then set up equations that represent the problem.

linear equations

Linear equations are mathematical statements representing relationships where each term is either a constant or the product of a constant and a single variable. In our problem, we can express the purchases of Troy and Lisa as linear equations. For Troy: 4 notebooks and 5 thumb drives cost \( 116 \) dollars, which we can write as:
\[ 4n + 5t = 116 \]
For Lisa: 2 notebooks and 3 thumb drives cost \( 68 \) dollars, which becomes:
\[ 2n + 3t = 68 \]
These equations form a system that we can solve using different methods.

elimination method

The elimination method allows us to remove one of the variables to solve the system of equations. Here’s how we apply it:

• First, we need to align the coefficients of one of the variables. Multiplying Equation 2 by 2, we get: \[ 4n + 6t = 136 \] (Equation 3)
• Next, we subtract Equation 1 from Equation 3 to eliminate \( n \):
  \[ (4n + 6t) - (4n + 5t) = 136 - 116 \]
  This simplifies to: \[ t = 20 \]
• With \( t \) found, we substitute \( t = 20 \) back into one of the original equations to find \( n \). Using Equation 2:
  \[ 2n + 3(20) = 68 \]
  \[ 2n + 60 = 68 \]
  \[ 2n = 8 \]
  \[ n = 4 \]

substitution method

Another method to solve systems of equations is the substitution method. It involves solving one equation for one variable and substituting this value into another equation. Here's an outline:

• Solve one equation for one variable. For example, solve Equation 2 for \( n \):
  \[ n = \frac{68 - 3t}{2} \]
• Substitute this expression into the other equation (Equation 1):
  \[ 4\bigg(\frac{68 - 3t}{2}\bigg) + 5t = 116 \]
  Simplify and solve for \( t \).
• Once you have \( t \), substitute it back into the equation \( n = \frac{68 - 3t}{2} \) to find \( n \).

While we used the elimination method in this problem, both methods are valuable tools for solving systems of linear equations.