Solving Y = -5x + 1010x And 2y = 20 Using Substitution A Comprehensive Guide

Introduction

In the realm of mathematics, solving systems of equations is a fundamental skill. These systems often represent real-world scenarios where multiple variables are interconnected. One powerful method for tackling these systems is the substitution method. In this comprehensive guide, we will delve into the process of using substitution to determine the number of solutions for the system of equations: y = -5x + 1010x and 2y = 20. We will explore each step meticulously, ensuring a clear understanding of the underlying principles and techniques involved. Whether you're a student grappling with algebra or simply seeking to refresh your knowledge, this article will provide you with the tools and insights needed to confidently solve such problems.

The ability to solve systems of equations is crucial in various fields, including engineering, economics, and computer science. The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. This allows us to substitute the expression into the other equation, thereby reducing the system to a single equation with a single variable. Let's embark on this mathematical journey and unravel the intricacies of this specific system of equations.

Understanding the Equations

Before we dive into the substitution method, it's crucial to understand the equations we're working with. Our system consists of two equations:

1. y = -5x + 1010x
2. 2y = 20

The first equation, y = -5x + 1010x, can be simplified by combining the 'x' terms. This simplification will make the substitution process more manageable. The second equation, 2y = 20, is straightforward and can be easily solved for 'y'. Let's simplify the first equation:

y = -5x + 1010x y = 1005x

Now our system of equations looks like this:

1. y = 1005x
2. 2y = 20

This simplified form makes it clearer how we can use the substitution method. The first equation expresses 'y' directly in terms of 'x', which is ideal for substitution. The second equation, as we'll see, allows us to find a specific value for 'y' quite easily. Understanding the structure of the equations is the first step towards solving the system effectively.

Applying the Substitution Method

Now that we have a clear understanding of our equations, we can proceed with the substitution method. The core idea behind this method is to solve one equation for one variable and then substitute that expression into the other equation. In our case, we already have the first equation solved for 'y':

y = 1005x

This makes our job easier. We can now substitute this expression for 'y' into the second equation:

2y = 20

Substituting y = 1005x into the second equation, we get:

2(1005x) = 20

This simplifies to:

2010x = 20

Now we have a single equation with one variable, 'x'. We can solve for 'x' by dividing both sides of the equation by 2010:

x = 20 / 2010 x = 2 / 201

So, we have found the value of 'x'. The next step is to use this value to find the value of 'y'. We can substitute the value of 'x' back into either of the original equations. The first equation, y = 1005x, is the most convenient for this purpose. Let's substitute x = 2 / 201 into the first equation:

y = 1005 * (2 / 201) y = 2010 / 201 y = 10

Thus, we have found the value of 'y'. The substitution method has led us to a unique solution for this system of equations.

Determining the Number of Solutions

After applying the substitution method, we found specific values for both 'x' and 'y'. This indicates that the system has a unique solution. In other words, there is only one pair of values (x, y) that satisfies both equations simultaneously. Let's recap the values we found:

• x = 2 / 201
• y = 10

This means the solution to the system of equations is the point (2/201, 10). To be absolutely sure, we can plug these values back into the original equations to verify that they hold true. Let's check:

1. y = -5x + 1010x 10 = -5(2/201) + 1010(2/201) 10 = -10/201 + 2020/201 10 = 2010/201 10 = 10 (True)
2. 2y = 20 2(10) = 20 20 = 20 (True)

Since the values satisfy both equations, we can confidently conclude that the system has one solution only. This solution is the point (2/201, 10). In general, a system of linear equations can have one solution, no solution, or infinitely many solutions. The substitution method helps us determine which of these scenarios applies to a given system.

Graphical Interpretation

To further solidify our understanding, let's consider the graphical interpretation of the system of equations. Each equation in the system represents a line in the coordinate plane. The solution to the system is the point where these lines intersect. In our case, since we found one unique solution, it means the two lines intersect at exactly one point.

The first equation, y = -5x + 1010x which simplifies to y = 1005x, represents a line that passes through the origin (0, 0) with a steep positive slope of 1005. The second equation, 2y = 20, simplifies to y = 10, which represents a horizontal line at y = 10. The intersection of these two lines is the point (2/201, 10), which confirms our algebraic solution.

If the lines were parallel, they would never intersect, and the system would have no solution. If the lines were the same, they would intersect at every point, and the system would have infinitely many solutions. The graphical perspective provides a visual way to understand the nature of the solutions to a system of equations.

Common Mistakes to Avoid

When solving systems of equations using the substitution method, it's easy to make mistakes if you're not careful. Here are some common pitfalls to watch out for:

1. Incorrect Substitution: Ensure you substitute the entire expression for the variable, not just a part of it. For example, if you have y = 2x + 3, make sure you replace 'y' with the entire expression 2x + 3 in the other equation.
2. Sign Errors: Be mindful of signs, especially when distributing or simplifying equations. A single sign error can lead to an incorrect solution.
3. Arithmetic Errors: Double-check your arithmetic calculations, such as multiplication, division, addition, and subtraction. Even a small error can throw off the entire solution.
4. Forgetting to Solve for Both Variables: Once you find the value of one variable, don't forget to substitute it back into one of the equations to find the value of the other variable. The solution to a system of equations consists of values for all variables involved.
5. Not Simplifying Equations: Before substituting, simplify the equations as much as possible. This can make the substitution process easier and reduce the chances of making mistakes.

By being aware of these common mistakes, you can increase your accuracy and efficiency when using the substitution method.

Conclusion

In this comprehensive guide, we have explored the substitution method for solving systems of equations. We tackled the specific system y = -5x + 1010x and 2y = 20, systematically walking through each step of the process. We simplified the equations, substituted one expression into another, solved for the variables, and verified our solution. We determined that the system has one solution only, which is the point (2/201, 10).

We also discussed the graphical interpretation of the system, which reinforced our understanding of how the solution represents the intersection of two lines. Furthermore, we highlighted common mistakes to avoid when using the substitution method, helping you to improve your problem-solving skills.

The substitution method is a versatile and powerful tool for solving systems of equations. By mastering this technique, you'll be well-equipped to tackle a wide range of mathematical problems in various contexts. Remember to practice regularly and pay attention to detail to ensure accuracy and efficiency. With a solid understanding of the substitution method, you can confidently approach and solve systems of equations, unlocking a crucial skill in the world of mathematics.