Problem 33 - Understanding Elementary Algebra with Geometry

Question

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. A mathematics department has budgeted \(\$ 10,000\) to purchase computers and printers. On this fixed budget they can purchase 10 computers and 10 printers, or they can purchase 12 computers and 2 printers. Find the individual costs of a computer and printer.

Step-by-Step Solution

Answer

The computer costs \(\$800\) and the printer costs \(\$200\).

Step 1: Define Variables

Let the cost of one computer be represented by \(c\) dollars and the cost of one printer be represented by \(p\) dollars.

Step 2: Set Up the First Equation

From the first scenario, they can purchase 10 computers and 10 printers for \(\$10,000\). Therefore, the equation is: \[10c + 10p = 10,000.\]

Step 3: Simplify First Equation

Divide the entire equation \(10c + 10p = 10,000\) by 10: \[c + p = 1,000.\]

Step 4: Set Up the Second Equation

From the second scenario, they can purchase 12 computers and 2 printers for \(\$10,000\). Therefore, the equation is: \[12c + 2p = 10,000.\]

Step 5: Solve the System of Equations

Using the simplified first equation \(c + p = 1,000\), solve for \(p\): \[p = 1,000 - c.\] Substitute \(p\) into the second equation: \[12c + 2(1,000 - c) = 10,000.\]

Step 6: Simplify and Solve for c

Expand and simplify: \[12c + 2,000 - 2c = 10,000,\] which simplifies to: \[10c + 2,000 = 10,000.\] Subtract 2,000 from both sides: \[10c = 8,000,\] then divide by 10: \[c = 800.\]

Step 7: Substitute to Find p

Substitute \(c = 800\) back into the equation \(p = 1,000 - c\): \[p = 1,000 - 800 = 200.\]

Step 8: State the Solution

The cost of one computer is \(\$800\) and the cost of one printer is \(\$200\).

Key Concepts

Solving Systems of Equations

In algebra, a system of equations involves two or more equations working together. The goal is to find a set of values that satisfies all equations in the system. In our example problem, we encounter two scenarios concerning the purchase of computers and printers. Each scenario provides a distinct equation based on the number of devices and the total budget. By solving these equations together, we can determine the cost of each item. The equations given in the problem are:

• 10 computers + 10 printers = \(10,000; and
• 12 computers + 2 printers = \)10,000.

These types of problems are ideal for practicing solving systems of equations, which is essential for many real-world applications.

Defining Variables

Defining variables is a crucial first step in transforming a word problem into an algebraic equation. In our problem, we are asked to find the costs of a computer and a printer. To do this, we assign variables:

• Let c be the cost of one computer, and
• let p be the cost of one printer.

Assigning these variables helps us structure the given information into mathematical equations. Understanding how to define variables effectively makes it easier to develop equations and eventually solve the problem.

Substitution Method

The substitution method is a common technique for solving systems of equations where one equation is solved for one variable, and then that expression is substituted into another equation. In our example:

• From the simplified first equation, we have:
  c + p = 1,000.
• We solve this for p:
  p = 1,000 - c.
• We then substitute p in the second equation:
  12c + 2(1,000 - c) = 10,000.

By substituting 1,000 - c for p, we transform the second equation into a single variable equation which we can solve for c. This method simplifies the system and leads us to the solution efficiently. Knowing substitution provides a powerful tool for solving more complex algebraic problems.

Budget Constraints

A budget constraint refers to the limit on the available spending of an entity, such as a mathematics department in our problem. This constraint means you must make choices within the budget allowed. In the exercise, we see two scenarios provided by the department's purchases:

• 10 computers and 10 printers cost \(10,000, and
• 12 computers and 2 printers also cost \)10,000.

Both scenarios represent different ways to allocate the $10,000 budget. By setting up equations based on these constraints, we reveal the individual costs of the computers and printers. Understanding how budget constraints work is essential for solving not only academic problems but also real-life financial decisions.