Problem 172 - Elementary Algebra

Question

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for \(\$ 40\) or he is able to buy 5 packages of paper and 6 staplers for \(\$ 62\). How much does a package of paper cost? How much does a stapler cost?

Step-by-Step Solution

Answer

A package of paper costs \(4, and a stapler costs \)7.

Step 1 - Define the variables

Let the cost of one package of paper be denoted by \( x \) dollars, and the cost of one stapler be \( y \) dollars.

Step 2 - Formulate the equations

Using the information given, we can write two equations: Equation 1: \( 3x + 4y = 40 \) Equation 2: \( 5x + 6y = 62 \)

Step 3 - Solve for one variable

We can use either substitution or elimination. Here, we use elimination. First, multiply Equation 1 by 3 and Equation 2 by 2 to align the coefficients of \( y \): \( 9x + 12y = 120 \) \( 10x + 12y = 124 \)

Step 4 - Eliminate one variable

Subtract Equation 1 from Equation 2 to eliminate \( y \): \( (10x + 12y) - (9x + 12y) = 124 - 120 \) This simplifies to: \( x = 4 \)

Step 5 - Solve for the remaining variable

Now substitute \( x = 4 \) back into either original equation (use Equation 1): \( 3(4) + 4y = 40 \) Simplifies to: \( 12 + 4y = 40 \) Then \( 4y = 28 \) So \( y = 7 \)

Step 6 - Verify the solution

Verify the solution by substituting \( x = 4 \) and \( y = 7 \) into both original equations: For Equation 1: \( 3(4) + 4(7) = 40 \) which is true. For Equation 2: \( 5(4) + 6(7) = 62 \) which is also true.

Key Concepts

linear equations

Peter's office supplies buying problem is a great example of a linear equation application. A linear equation is a mathematical expression that forms a straight line when plotted on a graph. Linear equations often take the form, \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants and \(x\) and \(y\) are variables.
In Peter's problem:

• 3 packages of paper and 4 staplers cost \(\$40\).
• 5 packages of paper and 6 staplers cost \(\$62\).

These can be expressed as two linear equations:

• \(3x + 4y = 40\)
• \(5x + 6y = 62\)

Linear equations help us model and solve real-world problems by relating different quantities.

elimination method

The elimination method is a way to solve a system of linear equations. It involves manipulating the equations to eliminate one variable, making it easier to solve for the other variable.
The steps for elimination are:

• 1. Align the equations and decide which variable to eliminate.
• 2. Multiply one or both equations to get the same coefficient for one variable in both equations.
• 3. Subtract one equation from the other to eliminate the chosen variable.
• 4. Solve the resulting equation for the remaining variable.

In Peter's problem, we eliminate variable \(y\) by:

• Multiplying the first equation by 3: \(9x + 12y = 120\).
• Multiplying the second equation by 2: \(10x + 12y = 124\).
• Subtracting the first from the second: \(x = 4\).

This simplifies our problem significantly, allowing us to then solve for \(x\).

substitution method

The substitution method is another way to solve linear equations. It involves solving one of the equations for one variable and substituting that expression into the second equation.
The steps for substitution are:

• 1. Solve one of the equations for one variable in terms of the other.
• 2. Substitute this expression into the other equation, reducing it to a single variable equation.
• 3. Solve the remaining equation for the single variable.
• 4. Substitute back to find the other variable.

In Peter's problem, if we use the substitution method, solve the first equation for \(x\) or \(y\):

• \(3x + 4y = 40\) can be written as \(x = \frac{40 - 4y}{3}\).
• Substitute \(x = \frac{40 - 4y}{3}\) into the second equation: \(5\left(\frac{40 - 4y}{3}\right) + 6y = 62\).
• This results in a single equation \(\frac{200 - 20y}{3} + 6y = 62\).

Solving this, we get ''y'' and then substitute back to find ''x''.

variables

In the world of mathematics, a variable is a symbol used to represent a number in equations and expressions. In Peter's problem, we defined:

• \(x\) as the cost of one package of paper.
• \(y\) as the cost of one stapler.

By defining these variables, we transformed a word problem into a solvable mathematical format.
Variables act as placeholders that allow us to set up our equations and manipulate them systematically. Always start by clearly defining your variables to ensure your equations accurately reflect the problem you're solving. Properly set variables lead to correct equations and, eventually, to correct solutions.