Understanding the System of Linear Equations

Oct 13, 2025 by miodeya.com 62 views

Introduction

When dealing with systems of linear equations, it is essential to determine the solution set, which can be one solution, no solutions, or infinitely many solutions. In this article, we will explore how to determine the solution set of a given system of linear equations. We will use the system of linear equations below as an example:

{ \begin{array}{l} 4x + 5y = 11 \\ 5y = -4x - 8 \end{array} \}

Understanding the System of Linear Equations

A system of linear equations is a set of two or more linear equations that involve the same set of variables. In this case, we have two linear equations with two variables, x and y. The first equation is 4x + 5y = 11, and the second equation is 5y = -4x - 8.

Solving the System of Linear Equations

To determine the solution set of the system of linear equations, we can use the method of substitution or elimination. In this case, we will use the method of substitution.

First, we will solve the second equation for y:

5y = -4x - 8

y = (-4x - 8) / 5

Now, we will substitute this expression for y into the first equation:

4x + 5y = 11

4x + 5((-4x - 8) / 5) = 11

Simplifying the Equation

To simplify the equation, we will multiply both sides by 5 to eliminate the fraction:

20x + 5(-4x - 8) = 55

Expanding the equation, we get:

20x - 20x - 40 = 55

Simplifying further, we get:

-40 = 55

This is a contradiction, as the left-hand side is negative and the right-hand side is positive.

Conclusion

Since we have a contradiction, we can conclude that the system of linear equations has no solutions.

Why No Solutions?

The reason why the system of linear equations has no solutions is that the two equations are inconsistent. The first equation is 4x + 5y = 11, and the second equation is 5y = -4x - 8. When we substitute the expression for y from the second equation into the first equation, we get a contradiction. This means that the two equations cannot be true at the same time, and therefore, the system of linear equations has no solutions.

Infinitely Many Solutions or No Solutions?

In some cases, a system of linear equations may have infinitely many solutions or no solutions. However, in this case, we have determined that the system of linear equations has no solutions.

Why Not Infinitely Many Solutions?

The reason why the system of linear equations does not have infinitely many solutions is that the two equations are inconsistent. When we substitute the expression for y from the second equation into the first equation, we get a contradiction. This means that the two equations cannot be true at the same time, and therefore, the system of linear equations has no solutions.

Conclusion

In conclusion, we have determined that the system of linear equations has no solutions. This is because the two equations are inconsistent, and when we substitute the expression for y from the second equation into the first equation, we get a contradiction.

Final Answer

The final answer is A. No Solution.

Discussion

This problem is a classic example of a system of linear equations with no solutions. The key to solving this problem is to recognize that the two equations are inconsistent and that the substitution method will lead to a contradiction.

Tips and Tricks

When dealing with systems of linear equations, it is essential to recognize that the two equations may be inconsistent. In this case, we can use the method of substitution to determine that the system of linear equations has no solutions.

Real-World Applications

Systems of linear equations have many real-world applications, including economics, physics, and engineering. In these fields, systems of linear equations are used to model real-world problems and to make predictions.

Conclusion

Final Answer

The final answer is A. No Solution.

References

[1] "Linear Algebra and Its Applications" by Gilbert Strang
[2] "Introduction to Linear Algebra" by Jim Hefferon
[3] "Linear Algebra" by David C. Lay

Additional Resources

[1] Khan Academy: Systems of Linear Equations
[2] MIT OpenCourseWare: Linear Algebra
[3] Wolfram Alpha: Systems of Linear Equations