Equation For Total Weight: 10 Lb & 3 Lb Weights

Oct 20, 2025 by ADMIN 48 views

Hey guys! Let's break down this math problem step-by-step. We've got a situation where Bob has both 10 lb and 3 lb weights, and all together they add up to a total of 50 lb. The goal here is to figure out an equation that helps us determine how many of each type of weight Bob has. We will use mathematical equations to represent real-world scenarios. This involves translating the given information into algebraic expressions and forming an equation.

Understanding the Problem

First, let’s clearly define our variables. If $x$ represents the number of 3 lb weights and $y$ represents the number of 10 lb weights, we need to create an equation that shows how these weights combine to reach 50 lb. The key here is to recognize that each type of weight contributes to the total, and we can express this contribution mathematically. Let's dive deeper into how we can construct this equation.

Breaking Down the Components

To start, consider how much the 3 lb weights contribute to the total. If Bob has $x$ number of these weights, then their combined weight is 3 multiplied by $x$ , or 3* $x$ . Similarly, if Bob has $y$ number of 10 lb weights, their combined weight is 10 multiplied by $y$ , or 10 $y$ *. The problem states that the total weight of all weights is 50 lb. This means we can add the combined weight of the 3 lb weights and the combined weight of the 10 lb weights to equal 50. This gives us the foundation for our mathematical equation.

Constructing the Equation

Now, let’s put it all together. The total weight from the 3 lb weights (3* $x$ ) plus the total weight from the 10 lb weights (10 $y$ *) must equal 50 lb. Mathematically, this is represented as:

3* $x$ * + 10* $y$ * = 50

This equation is the answer we're looking for. It represents the relationship between the number of 3 lb weights ( $x$ ) and the number of 10 lb weights ( $y$ ) that Bob has, given that their total weight is 50 lb. This equation allows us to explore different combinations of $x$ and $y$ that satisfy the condition. Understanding how to set up such equations is crucial in solving a variety of mathematical and real-world problems.

Why This Equation Works

The reason this equation works so well is that it precisely translates the word problem into a mathematical statement. It captures the essence of the relationship between the quantities involved. Each term in the equation represents a specific component of the total weight:

3* $x$ *: Represents the total weight contributed by the 3 lb weights.
10* $y$ *: Represents the total weight contributed by the 10 lb weights.
50: Represents the overall total weight.

By setting the sum of the individual contributions equal to the total, we create a model that accurately reflects the situation described in the problem. This model can then be used to solve for unknowns or explore different scenarios. For example, if we knew the number of 3 lb weights, we could substitute that value for $x$ and solve for $y$ , the number of 10 lb weights. This is the power of using equations to model real-world situations.

Solving the Equation (Further Exploration)

While the question specifically asks for the equation, let’s briefly touch on how we might solve it to find possible values for $x$ and $y$ . The equation 3* $x$ * + 10* $y$ * = 50 is a linear Diophantine equation, which means we're looking for integer solutions (since Bob can't have a fraction of a weight). To solve this type of equation, we typically look for one solution first, and then find a general solution. However, let's focus on finding possible solutions within the context of the problem.

Finding Possible Solutions

Since $x$ and $y$ represent the number of weights, they must be non-negative integers (0, 1, 2, 3, ...). We can try different values for one variable and see if we get an integer value for the other. For instance:

If $y$ = 0 (no 10 lb weights), then 3* $x$ * = 50. Solving for $x$ , we get $x$ = 50/3, which is not an integer.
If $y$ = 1 (one 10 lb weight), then 3* $x$ * + 10 = 50, so 3* $x$ * = 40. Again, $x$ = 40/3 is not an integer.
If $y$ = 2 (two 10 lb weights), then 3* $x$ * + 20 = 50, so 3* $x$ * = 30, which gives $x$ = 10. This is an integer solution!
If $y$ = 3 (three 10 lb weights), then 3* $x$ * + 30 = 50, so 3* $x$ * = 20, and $x$ = 20/3, not an integer.
If $y$ = 4 (four 10 lb weights), then 3* $x$ * + 40 = 50, so 3* $x$ * = 10, and $x$ = 10/3, not an integer.
If $y$ = 5 (five 10 lb weights), then 3* $x$ * + 50 = 50, so 3* $x$ * = 0, which gives $x$ = 0. This is another integer solution!

So, we’ve found two possible solutions: Bob could have 10 three-pound weights and 2 ten-pound weights, or he could have 0 three-pound weights and 5 ten-pound weights. This exploration highlights how a single equation can sometimes have multiple solutions, especially when dealing with real-world constraints.

Importance of Defining Variables

In this problem, defining variables clearly is super important. When we say $x$ represents the number of 3 lb weights and $y$ represents the number of 10 lb weights, we're setting the foundation for translating the word problem into mathematical language. Without this clear definition, it would be challenging to construct a meaningful equation. Defining variables helps us:

Organize Information: It allows us to keep track of what each symbol represents.
Translate Words to Math: It provides a direct link between the problem statement and the equation.
Avoid Confusion: It reduces ambiguity and ensures everyone understands what we're talking about.
Solve Systematically: It guides us in setting up the equation and finding solutions.

Think of variables as the building blocks of our mathematical model. They are the containers that hold the quantities we're interested in, and defining them precisely is the first step in solving the problem.

Real-World Applications

The ability to create and solve equations like this is useful in many real-world situations. Whether you're figuring out ingredient quantities for a recipe, managing a budget, or planning a construction project, mathematical equations can help you make informed decisions. This specific type of problem, involving integer solutions, comes up in various scenarios, such as:

Inventory Management: Determining how many of each item to order given budget constraints.
Currency Exchange: Figuring out how many bills of different denominations to use for a certain amount.
Resource Allocation: Deciding how to distribute resources to different tasks or projects.

The key takeaway here is that math isn't just about abstract concepts; it's a powerful tool for problem-solving in everyday life. By understanding how to translate real-world situations into mathematical models, we can tackle a wide range of challenges more effectively.

Conclusion

So, to wrap things up, the equation that can be used to find the number of each type of weight Bob has is:

3* $x$ * + 10* $y$ * = 50

This equation beautifully captures the relationship between the 3 lb weights and 10 lb weights, allowing us to explore different combinations that add up to 50 lb. Remember, the power of mathematics lies in its ability to model and solve real-world problems. By practicing these skills, you'll be well-equipped to tackle a wide range of challenges! Keep up the great work, guys!