Determining Solutions For Linear Systems A Step-by-Step Guide

In mathematics, a linear system, also known as a system of linear equations, is a collection of two or more linear equations involving the same set of variables. A solution to a linear system is a set of values for the variables that satisfies all equations in the system simultaneously. Understanding how many solutions a linear system possesses is a fundamental concept in linear algebra. Linear systems can have one solution, no solution, or an infinite number of solutions. This article delves into the methods for determining the number of solutions a given linear system has, using the example provided:

$ \begin{array}{l} y=2 x-5 \ -8 x-4 y=-20 \end{array} $

We will explore different approaches, including substitution, elimination, and graphical methods, to classify the system and find its solution set. Let's begin by understanding the basics of linear systems and their solutions.

Understanding Linear Systems

A linear system consists of two or more linear equations. A linear equation is an equation in which the highest power of the variables is 1. For example, $y = 2x - 5$ and $-8x - 4y = -20$ are both linear equations. When we have a system of such equations, we seek values for the variables (in this case, $x$ and $y$) that satisfy all equations simultaneously.

There are three possible outcomes when solving a linear system:

1. One Solution: The lines intersect at a single point. This means there is exactly one pair of $(x, y)$ values that satisfy both equations.
2. No Solution: The lines are parallel and never intersect. This indicates that there are no $(x, y)$ values that can satisfy both equations simultaneously.
3. Infinite Solutions: The lines are coincident, meaning they are the same line. Every point on the line represents a solution, leading to an infinite number of solutions.

Methods for Solving Linear Systems

To determine the number of solutions, we can use several methods:

• Substitution: Solve one equation for one variable and substitute that expression into the other equation.
• Elimination: Add or subtract the equations to eliminate one variable.
• Graphical Method: Plot the equations on a coordinate plane and observe their intersection.

Solving the Given Linear System

Let's apply these methods to the given system:

$ \begin{array}{l} y=2 x-5 \ -8 x-4 y=-20 \end{array} $

Method 1: Substitution

The first equation is already solved for $y$, so we can substitute this expression into the second equation:

$-8x - 4(2x - 5) = -20$

Now, we simplify and solve for $x$:

$-8x - 8x + 20 = -20$

$-16x = -40$

$x = \frac{-40}{-16} = \frac{5}{2} = 2.5$

Now that we have the value of $x$, we can substitute it back into the first equation to find $y$:

$y = 2(2.5) - 5$

$y = 5 - 5 = 0$

So, the solution is $(2.5, 0)$. This suggests that the system has one solution. However, we should also check the elimination method to confirm.

Method 2: Elimination

To use the elimination method, we need to align the variables in both equations. The given system is:

$ \begin{array}{l} y=2 x-5 \ -8 x-4 y=-20 \end{array} $

Rearrange the first equation to align the variables:

$-2x + y = -5$

Now we have:

$ \begin{array}{l} -2x + y = -5 \ -8 x-4 y=-20 \end{array} $

To eliminate $x$, we can multiply the first equation by -4:

$-4(-2x + y) = -4(-5)$

$8x - 4y = 20$

Now add the modified first equation to the second equation:

$ \begin{array}{l} 8x - 4y = 20 \ -8 x-4 y=-20 \end{array} $

Adding the equations, we get:

$(8x - 8x) + (-4y - 4y) = 20 - 20$

$0 - 8y = 0$

$-8y = 0$

$y = 0$

Substitute $y = 0$ into the first equation $-2x + y = -5$:

$-2x + 0 = -5$

$-2x = -5$

$x = \frac{5}{2} = 2.5$

Again, we find the solution $(2.5, 0)$, confirming that there is one solution.

Method 3: Graphical Method

To use the graphical method, we plot both equations on a coordinate plane. The equations are:

$y = 2x - 5$

$-8x - 4y = -20$

First, rewrite the second equation in slope-intercept form ($y = mx + b$):

$-4y = 8x - 20$

$y = -2x + 5$

Now we have:

$ \begin{array}{l} y = 2x - 5 \ y = -2x + 5 \end{array} $

Plotting these two lines, we see that they intersect at the point $(2.5, 0)$. This graphical representation visually confirms that the system has one solution. The intersection point represents the unique solution to the system.

Analyzing the Results

Through the substitution, elimination, and graphical methods, we have consistently found that the given linear system has one solution, which is $(2.5, 0)$. This means that the two lines represented by the equations intersect at a single point on the coordinate plane. When we solved the system algebraically using both substitution and elimination, we arrived at the same solution, providing further confirmation of our result. The graphical method visually demonstrated the intersection of the two lines at the point $(2.5, 0)$, reinforcing our conclusion.

Verifying the Solution

To ensure our solution is correct, we can substitute the values $x = 2.5$ and $y = 0$ back into the original equations:

Equation 1: $y = 2x - 5$

$0 = 2(2.5) - 5$

$0 = 5 - 5$

$0 = 0$ (True)

Equation 2: $-8x - 4y = -20$

$-8(2.5) - 4(0) = -20$

$-20 - 0 = -20$

$-20 = -20$ (True)

Since the values satisfy both equations, we can confidently say that $(2.5, 0)$ is the correct solution.

Conclusion

In summary, the linear system

$ \begin{array}{l} y=2 x-5 \ -8 x-4 y=-20 \end{array} $

has one solution, which is the point $(2.5, 0)$. We arrived at this conclusion by employing three different methods: substitution, elimination, and graphical representation. Each method independently led us to the same solution, and we verified the solution by substituting the values back into the original equations. Understanding the different methods for solving linear systems and interpreting the results is crucial in algebra and has broad applications in various fields, including engineering, economics, and computer science. Knowing how to determine whether a system has one solution, no solution, or infinite solutions allows us to model and solve real-world problems effectively. By mastering these techniques, students and professionals can approach complex problems with confidence and accuracy.

This exercise underscores the importance of a multifaceted approach to problem-solving in mathematics. Utilizing multiple methods not only confirms the accuracy of the solution but also deepens our understanding of the underlying concepts. The ability to solve linear systems is a foundational skill that opens the door to more advanced mathematical topics and their practical applications.

The correct answer is B. one solution: (2.5,0).