Problem 71 - Elementary Algebra

Question

Find the solution. This year an item costs \(\$ 106\), an increase of \(\$ 10\) over last year's price. What was last year's price?

Step-by-Step Solution

Answer

Answer: The price of the item last year was $96.

Step 1: Identify the information given

We are given that the current price of the item is \(\$ 106\), and that this is an increase of \(\$ 10\) from last year's price. We want to find last year's price.

Step 2: Set up the equation

Let's denote last year's price as \(P\). Since the current price increased by \(\$ 10\), we can set up an equation: $$106 = P + 10$$

Step 3: Solve for last year's price (P)

To find the value of \(P\), we will need to subtract \(10\) from both sides of the equation: $$106 - 10 = P$$.$$96 = P$$

Step 4: State the final answer

Last year's price was \(\$ 96\).

Key Concepts

Problem Solving

Problem-solving is a fundamental skill in mathematics that allows us to tackle a wide variety of situations by identifying what is known and what needs to be found. In this exercise, we need to find last year's price of an item based on its current price and the increase in price over the past year.

First, we start by understanding the data given in the problem. We know two key pieces of information:

• The current price of the item is \\( 106\.
• This price represents an increase of \\) 10\ over last year's price.

With these pieces of information, we set up a plan: we will use an algebraic equation to express the relationship between last year's price and this year's price.

Solving real-world problems using this method involves critically thinking about what you need to find and using the given information effectively.

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. They can often be expressed in the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants.

In this exercise, we have used a linear equation to find the price of an item from the previous year. The equation \(106 = P + 10\) represents the relationship between the current price and last year's price:

• \(P\) is the variable representing last year's price.
• The number \10\ is added to \(P\) to account for this year's increase.
• The result of this expression is the current price, which is \106\.

Understanding linear equations helps us model situations where one quantity changes in relation to another, providing a straightforward path to finding unknown values.

Basic Algebra Concepts

Basic algebra concepts are foundational to understanding how to solve equations and find unknowns. One key concept is solving equations by isolating the variable, which involves performing the same operation on both sides of the equation until the variable is by itself.

In our exercise, the goal was to isolate \(P\) to find last year's price. We did this by performing the same subtraction operation on both sides of the equation, subtracting 10 from 106 to find that \(P = 96\).

• Algebra efficiently organizes and solves the problem by representing unknown quantities with variables.
• It allows us to structure and solve equations systematically.
• Practicing these concepts builds confidence and skill in handling more complex algebraic problems in the future.

Mastering basic algebraic techniques ensures that you can approach similar problems with a clear and logical method.