Problem 63 - Algebra for College Students

Question

Use an algebraic approach to solve each problem. The selling price of a ring is \(\$ 750\). This represents \(\$ 150\) less than three times the cost of the ring. Find the cost of the ring.

Step-by-Step Solution

Answer

The cost of the ring is $300.

Step 1: Define the variables

Let \( x \) be the cost of the ring. This is the unknown value we are trying to find.

Step 2: Set up the equation

According to the problem, the selling price of the ring is equal to \( \\(750 \), and it represents \( \\)150 \) less than three times the cost of the ring. This can be written as the equation: \[ 750 = 3x - 150 \]

Step 3: Solve for the cost of the ring

To find \( x \), start by adding \( 150 \) to both sides to isolate the term containing \( x \) on one side of the equation:\[ 750 + 150 = 3x \]\[ 900 = 3x \]

Step 4: Divide by the coefficient of x

To solve for \( x \), divide both sides of the equation by 3:\[ x = \frac{900}{3} \]\[ x = 300 \]

Key Concepts

Cost Analysis

Cost analysis helps us break down the expenses involved in purchasing an item to understand the true cost behind it. In the context of this ring problem, we are evaluating the cost by comparing it to its selling price. The selling price is often set higher than the cost to ensure profitability. However, in certain questions like our ring scenario, we have specific conditions that reverse-engineer the original price: we know the selling price and an algebraic relation concerning the cost.

To perform cost analysis, we need to consider:

• The known selling price
• The equation that relates cost and selling price (in this case, "three times the cost minus $150" as expressed in the problem statement)

Pinpointing the initial cost involves understanding the algebraic relationship that affects the final selling price, allowing us to backtrack to the original expense.

Linear Equations

Linear equations form the backbone of many algebraic problems, allowing us to model relationships between variables. In this exercise, we are given a linear equation: \[750 = 3x - 150\]

In simple terms, a linear equation represents a straight line when graphed. Here, the relationship between the selling price and the cost of the ring is linear because it involves a constant rate ("three times") and a constant offset ("less 150").

To solve a linear equation, our goal is to isolate the variable in question ("x" for cost) on one side of the equation. This involves a few straightforward steps such as:

• Adding or subtracting terms to both sides to maintain the equation's balance
• Dividing or multiplying to simplify the solution

The beauty of linear equations lies in their simplicity and direct approach to finding unknown values.

Problem-Solving Techniques

When tackling algebraic problems, employing effective problem-solving techniques can simplify the process significantly. Here are a few methods highlighted in our ring cost problem:

First, **define the variable**. By clearly identifying what you need to find, you set a clear goal. Here, that goal is the cost of the ring, denoted by the variable "\( x \)".

Next, **set up the equation**. Using the relationships described in the word problem, translate them into mathematical equations. In our example, this is reflected as: \[750 = 3x - 150\]This step involves careful reading and comprehension of the problem.

Following this, **solve the equation** using algebraic manipulation. Our steps include adding 150 to both sides and then dividing by 3, simplifying towards the solution "\( x = 300 \)".

Finally, **verify your result** by checking that it satisfies the original conditions of the problem. This ensures the process was correct and enhances your understanding of problem-solving as a whole. By taking these methods into account, you make solving algebraic equations more structured and less daunting.