Question

A student sold a computer and a printer for a total of \(\$ 995,\) getting \(1 \frac{1}{2}\) times as much for the printer as for the computer. What was the price of each?

Short answer

The computer was sold for \$398 and the printer for \$597.

Step-by-step solution

Define Variables

Let the price of the computer be represented by the variable 'c'. Since the printer was sold for one and a half times the amount of the computer, the price of the printer can be represented as '1.5c'.

Write the Equation

Using the information given, write an equation that represents the sum of the prices of the computer and the printer which equals $995. The equation will be: \( c + 1.5c = 995 \).

Combine Like Terms

Combine the 'c' terms to simplify the equation: \( 1c + 1.5c = 2.5c \), so the equation now is \( 2.5c = 995 \).

Solve for 'c'

Divide both sides of the equation by 2.5 to solve for 'c': \( c = \frac{995}{2.5} \). Calculate the result to find the price of the computer.

Find the Price of the Printer

Now that we know 'c', calculate the price of the printer by multiplying 'c' by 1.5 since the printer is sold for one and a half times the price of the computer. The price of the printer is \(1.5c\).

Key concepts

Defining Variables in Algebra

In algebra, defining variables is a fundamental first step in translating word problems into mathematical equations. It's the process of choosing a letter or symbol to represent an unknown quantity in a problem. Imagine you're a detective, and your case is to find out the value of this unknown. By assigning a variable, like 'c' for the cost of a computer, you are naming your suspect, which then helps you solve the case.

Variables should represent the quantities in a problem and make sense in context. For instance, if we're dealing with the price of a computer and a printer, using 'c' for the computer and 'pc' or '1.5c' for the printer provides clarity and simplicity when forming equations. Choosing appropriate variables not only helps in simplifying the problem but also in avoiding confusion later on when solving the equations.

Writing Algebraic Equations

After variables are defined, writing algebraic equations is like scripting a play where the variables are characters and the numbers tell the story of their relationships. To write an algebraic equation, you translate the problem's written statements into mathematical language, using your variables along with the appropriate numbers and operations.

In our example, the total sale price of a computer and a printer adds up to \(995. So, we craft an equation to match: that the price of the computer 'c' plus one and a half times 'c' for the printer equals \)995. It's a simple yet elegant way to capture all the information given to us in a language math understands: an equation.

Combining Like Terms

When we have an algebraic expression with similar terms, combining them simplifies the problem. Like terms are the detective's clues that fit together because they have the same variable raised to the same power. When we combine them, we're adding or subtracting their coefficients — the numbers in front of the variables.

Think of it as a team of workers where each one represents a term with 'c' in our equation. To combine the workload efficiently, we group the same types of workers together. In our case, 'c' and '1.5c' are on the same team, so we add them up to get '2.5c', leaving us with a simpler equation and a clearer path to the solution.

Solving Linear Equations

The final act in the algebra drama is solving linear equations. It's when you discover the true identity of the variable you've been tracking. For linear equations, our goal is to isolate the variable on one side of the equation. This means performing operations that 'undo' what's being done to the variable.

In our story, we have '2.5c = 995', and we want to find the value of 'c' alone. By dividing both sides of the equation by 2.5, we're effectively peeling away what's been added to 'c', revealing its value. It's like finding the key that unlocks the mystery of the problem, providing us the price of the computer and leading us one step closer to knowing the price of the printer as well.