Solving Systems Of Equations A Step-by-Step Guide

In mathematics, a system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that make all the equations true simultaneously. Solving systems of equations is a fundamental skill in algebra and has applications in various fields, including engineering, economics, and computer science. In this article, we will delve into the process of solving a system of linear equations, focusing on the specific example provided:

\left\{\begin{array}{r} x+3 y=-9 \\ 6 x+4 y=-40 \end{array}\right.

We will explore the methods used to determine the solution and, in cases where the system is dependent, learn how to express the solution set in terms of one variable. Understanding these techniques is crucial for anyone looking to master algebra and its applications.

Understanding Systems of Equations

At its core, a system of equations represents a set of mathematical relationships between variables. Each equation in the system provides a constraint on the values that the variables can take. The goal of solving the system is to find the values that satisfy all the constraints simultaneously. Geometrically, each equation in a two-variable system represents a line on a coordinate plane. The solution to the system corresponds to the point(s) where the lines intersect. There are three possible scenarios:

1. Unique Solution: The lines intersect at a single point, indicating a unique solution.
2. No Solution: The lines are parallel and do not intersect, indicating that there is no solution that satisfies both equations.
3. Infinitely Many Solutions: The lines are coincident (i.e., they are the same line), indicating that there are infinitely many solutions. This is known as a dependent system.

The system we are tackling, consisting of the equations $x + 3y = -9$ and $6x + 4y = -40$, falls into one of these categories. To determine which one, we'll employ algebraic techniques to find the solution or demonstrate the nature of the system.

Methods for Solving Systems of Equations

There are several methods for solving systems of equations, including:

• Substitution: Solve one equation for one variable and substitute that expression into the other equation.
• Elimination (or Addition): Multiply one or both equations by constants to make the coefficients of one variable opposites, then add the equations together to eliminate that variable.
• Graphing: Graph both equations on the same coordinate plane and find the point(s) of intersection.
• Matrix Methods: Use matrices and matrix operations to solve the system (more suitable for larger systems).

For the given system, we will primarily use the elimination method, as it's often the most efficient for linear systems. However, we may also touch upon substitution to illustrate its applicability.

Solving the System Using Elimination

To solve the system

\left\{\begin{array}{r} x+3 y=-9 \\ 6 x+4 y=-40 \end{array}\right.

using elimination, our goal is to eliminate one of the variables by manipulating the equations. Let's eliminate $x$. To do this, we can multiply the first equation by -6, which will make the coefficient of $x$ in the first equation the opposite of the coefficient of $x$ in the second equation.

Multiplying the first equation by -6, we get:

-6(x + 3y) = -6(-9)

-6x - 18y = 54

Now we have the modified system:

\left\{\begin{array}{r} -6x - 18y = 54 \\ 6x + 4y = -40 \end{array}\right.

Adding the two equations together, we get:

(-6x - 18y) + (6x + 4y) = 54 + (-40)

-14y = 14

Dividing both sides by -14, we find:

y = -1

Now that we have the value of $y$, we can substitute it back into either of the original equations to solve for $x$. Let's use the first equation:

x + 3y = -9

Substitute $y = -1$:

x + 3(-1) = -9

x - 3 = -9

Adding 3 to both sides, we get:

x = -6

Thus, the solution to the system is $x = -6$ and $y = -1$. This means the point of intersection of the two lines represented by the equations is $(-6, -1)$.

Verifying the Solution

To ensure our solution is correct, we should substitute the values of $x$ and $y$ back into both original equations and verify that they hold true.

For the first equation, $x + 3y = -9$:

(-6) + 3(-1) = -6 - 3 = -9

This is true.

For the second equation, $6x + 4y = -40$:

6(-6) + 4(-1) = -36 - 4 = -40

This is also true. Therefore, our solution $x = -6$ and $y = -1$ is correct.

Expressing the Solution Set

Since we found a unique solution, the solution set consists of a single ordered pair. The solution set can be expressed as:

\{(-6, -1)\}

This indicates that there is only one point that satisfies both equations in the system.

Dependent Systems and Infinite Solutions

In some cases, a system of equations may be dependent, meaning that the equations represent the same line or multiples of each other. In such cases, there are infinitely many solutions. When solving a dependent system, you will typically encounter a situation where one equation reduces to an identity (e.g., $0 = 0$) after applying elimination or substitution.

To express the solution set for a dependent system, we express one variable in terms of the other. For example, if we had a system that simplified to $x + y = 2$, we could express the solution set as:

\{(x, y) | y = 2 - x\}

This notation indicates that for any value of $x$, the corresponding value of $y$ is $2 - x$, and the pair $(x, y)$ is a solution to the system.

Solving by Substitution (Alternative Method)

While we primarily used elimination, let's briefly illustrate how substitution could be used to solve the same system.

Starting with the system:

\left\{\begin{array}{r} x+3 y=-9 \\ 6 x+4 y=-40 \end{array}\right.

We can solve the first equation for $x$:

x = -9 - 3y

Now, substitute this expression for $x$ into the second equation:

6(-9 - 3y) + 4y = -40

-54 - 18y + 4y = -40

-14y = 14

y = -1

Now substitute $y = -1$ back into the expression for $x$:

x = -9 - 3(-1)

x = -9 + 3

x = -6

As we can see, the substitution method yields the same solution, $x = -6$ and $y = -1$.

Conclusion

In this comprehensive guide, we have successfully solved the system of equations

\left\{\begin{array}{r} x+3 y=-9 \\ 6 x+4 y=-40 \end{array}\right.

using the elimination method and verified the solution using substitution. We found a unique solution, $x = -6$ and $y = -1$, which we expressed as the solution set ${(-6, -1)}$. We also discussed the concept of dependent systems and how to express their solution sets in terms of one variable.

Mastering the techniques for solving systems of equations is crucial for success in algebra and its applications. By understanding the different methods and practicing regularly, you can confidently tackle a wide range of problems involving systems of equations.