Solving Linear Equations Find Ordered Pair Solution (m, N)
When faced with a system of linear equations, the goal is to find the values of the variables that satisfy all equations simultaneously. In other words, we seek the ordered pair (m, n) that, when substituted into both equations, makes both equations true. This article delves into the process of solving the given system of linear equations to determine the correct ordered pair solution. We will explore the concepts, techniques, and step-by-step methods to arrive at the accurate answer. Understanding how to solve systems of equations is a fundamental skill in algebra and has wide applications in various fields, making it an essential topic for students and professionals alike.
Understanding Systems of Linear Equations
A system of linear equations consists of two or more linear equations involving the same variables. A linear equation is an equation that can be written in the form Ax + By = C, where A, B, and C are constants, and x and y are variables. In a system of two linear equations with two variables, the solution is an ordered pair (x, y) that satisfies both equations. Graphically, the solution represents the point where the lines corresponding to the equations intersect. There are several methods to solve systems of linear equations, including substitution, elimination, and graphing. Each method has its advantages and may be more suitable for specific types of systems. The ability to solve these systems is crucial for modeling and solving real-world problems, making it a vital skill in mathematics and various applications.
Methods for Solving Linear Systems
There are primarily two algebraic methods for solving systems of linear equations: substitution and elimination. The substitution method involves solving one equation for one variable and substituting that expression into the other equation. This results in a single equation with one variable, which can be solved directly. Then, the value of that variable is substituted back into one of the original equations to find the value of the other variable. The elimination method, on the other hand, involves manipulating the equations so that the coefficients of one variable are opposites. By adding the equations, one variable is eliminated, leaving a single equation with one variable. This method is particularly effective when the coefficients of one variable are easily made opposites. Additionally, graphing provides a visual approach to solving systems. By graphing both equations on the same coordinate plane, the solution is the point of intersection of the lines. While graphing can be useful for understanding the concept and estimating the solution, algebraic methods are generally more accurate and efficient for complex systems. Choosing the appropriate method depends on the specific equations in the system and the solver's preference.
Given System of Equations
In this problem, we are given the following system of linear equations:
6m - 5n = 14
2m + 2n = -10
Our goal is to find the ordered pair (m, n) that satisfies both equations. We will explore both the substitution and elimination methods to determine the most efficient approach for solving this particular system. By understanding the structure of the equations and the relationships between the variables, we can strategically apply the appropriate method to arrive at the solution. Careful execution of each step is crucial to ensure accuracy and avoid common algebraic errors. The solution to this system will provide the values of m and n that make both equations true simultaneously, representing the point of intersection if these equations were graphed as lines.
Solving by Elimination Method
To solve the system using the elimination method, we first aim to eliminate one of the variables. Looking at the equations:
6m - 5n = 14
2m + 2n = -10
We can eliminate 'm' by multiplying the second equation by -3. This will make the coefficient of 'm' in the second equation -6, which is the opposite of the coefficient of 'm' in the first equation. Multiplying the second equation by -3, we get:
-3(2m + 2n) = -3(-10)
-6m - 6n = 30
Now we have the modified system:
6m - 5n = 14
-6m - 6n = 30
Adding the two equations eliminates 'm':
(6m - 5n) + (-6m - 6n) = 14 + 30
-11n = 44
Dividing both sides by -11, we find:
n = -4
Now that we have the value of n, we can substitute it back into one of the original equations to find the value of m. Let's use the second equation:
2m + 2n = -10
2m + 2(-4) = -10
2m - 8 = -10
Adding 8 to both sides:
2m = -2
Dividing by 2:
m = -1
Thus, the ordered pair solution is (-1, -4). This methodical approach ensures we systematically eliminate one variable and solve for the other, leading us to the correct solution.
Solving by Substitution Method
Alternatively, we can solve the system using the substitution method. This method involves solving one equation for one variable and substituting that expression into the other equation. Let's start with the second equation:
2m + 2n = -10
We can solve for m by first subtracting 2n from both sides:
2m = -10 - 2n
Then, divide by 2:
m = -5 - n
Now, substitute this expression for m into the first equation:
6m - 5n = 14
6(-5 - n) - 5n = 14
Distribute the 6:
-30 - 6n - 5n = 14
Combine like terms:
-11n = 44
Divide by -11:
n = -4
Substitute n = -4 back into the expression for m:
m = -5 - n
m = -5 - (-4)
m = -5 + 4
m = -1
Again, we find that the ordered pair solution is (-1, -4). The substitution method provides a different pathway to the same solution, reinforcing the correctness of our answer.
Verifying the Solution
To ensure the accuracy of our solution, it is crucial to verify the ordered pair (-1, -4) by substituting it back into both original equations. Let's start with the first equation:
6m - 5n = 14
Substitute m = -1 and n = -4:
6(-1) - 5(-4) = 14
-6 + 20 = 14
14 = 14
The first equation is satisfied. Now, let's check the second equation:
2m + 2n = -10
Substitute m = -1 and n = -4:
2(-1) + 2(-4) = -10
-2 - 8 = -10
-10 = -10
The second equation is also satisfied. Since the ordered pair (-1, -4) satisfies both equations, we have verified that it is indeed the correct solution to the system of linear equations. This step is an essential part of the problem-solving process, as it helps to catch any potential errors and ensures the reliability of the final answer.
Conclusion
In conclusion, we have successfully solved the given system of linear equations using both the elimination and substitution methods. The ordered pair that satisfies the system
6m - 5n = 14
2m + 2n = -10
is (-1, -4). We verified this solution by substituting the values of m and n back into both original equations, confirming that they hold true. This problem illustrates the importance of understanding and applying different methods for solving systems of linear equations. Whether using elimination, substitution, or other techniques, the key is to carefully execute each step and verify the solution to ensure accuracy. Mastering these skills is fundamental for success in algebra and beyond, as systems of equations arise in numerous applications across mathematics, science, and engineering. The ability to confidently solve these systems equips individuals with a powerful tool for problem-solving and analytical thinking.
