Solving Linear Equations X+y=4 And X-y=6 A Step-by-Step Guide
Systems of linear equations are a fundamental concept in algebra, representing a set of two or more linear equations with the same variables. The solution to a system of linear equations is the set of values for the variables that satisfy all equations simultaneously. In simpler terms, it's the point where the lines represented by the equations intersect on a graph. Understanding how to solve these systems is crucial for various applications in mathematics, science, engineering, and economics. This article delves into a common method for solving systems of linear equations, providing a clear, step-by-step approach to find the solution.
Methods for Solving Systems of Linear Equations
Several methods exist for solving systems of linear equations, including graphing, substitution, and elimination. Each method has its advantages and disadvantages, and the best choice often depends on the specific system of equations being solved. The graphing method, while visually intuitive, can be less accurate for non-integer solutions. The substitution method involves solving one equation for one variable and substituting that expression into the other equation, which is effective when one variable is easily isolated. However, the elimination method, also known as the addition or subtraction method, is particularly useful when the coefficients of one variable are opposites or can be easily made opposites. This method involves manipulating the equations to eliminate one variable, making it easier to solve for the remaining variable. This article will focus on the elimination method as a robust and efficient technique for solving systems of linear equations.
Elimination Method: A Detailed Walkthrough
The elimination method hinges on the principle of adding or subtracting equations to eliminate one variable. This simplification allows us to solve for the remaining variable and then back-substitute to find the value of the eliminated variable. The key steps involved in this method are:
- Align the equations: Ensure that like terms (terms with the same variable) are aligned vertically in the system of equations. This means that the x-terms, y-terms, and constant terms should be in the same columns. Proper alignment is crucial for the subsequent steps to work correctly.
- Make the coefficients of one variable opposites: Examine the coefficients of either x or y in both equations. If they are not opposites (e.g., 2 and -2), multiply one or both equations by a constant so that the coefficients of one variable become opposites. This step prepares the equations for the elimination process. For instance, if you have equations with 2x and 3x, you might multiply the first equation by 3 and the second by -2 to get 6x and -6x, respectively.
- Add the equations: Add the two equations together. The variable with opposite coefficients should cancel out, leaving you with a single equation in one variable. This step is the heart of the elimination method, where the system is simplified to a solvable form.
- Solve for the remaining variable: Solve the resulting equation for the remaining variable. This will give you the numerical value of one of the variables in the system.
- Substitute back: Substitute the value you found in the previous step into either of the original equations and solve for the other variable. This step completes the solution process by finding the value of the variable that was initially eliminated.
- Check the solution: Substitute both values back into the original equations to verify that they satisfy both equations. This is a crucial step to ensure the accuracy of your solution and catch any potential errors made during the process.
Solving the System: x + y = 4 and x - y = 6
Let's apply the elimination method to the given system of linear equations:
x + y = 4
x - y = 6
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Align the equations: The equations are already aligned, with x-terms, y-terms, and constants in their respective columns.
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Make the coefficients of one variable opposites: Notice that the coefficients of y are already opposites (+1 and -1). This simplifies our work significantly, as we don't need to multiply the equations by any constants.
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Add the equations: Add the two equations together:
(x + y) + (x - y) = 4 + 6 2x = 10
The y terms cancel out, leaving us with an equation in x.
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Solve for the remaining variable: Divide both sides of the equation by 2 to solve for x:
2x / 2 = 10 / 2 x = 5
We have found that x = 5.
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Substitute back: Substitute x = 5 into either of the original equations. Let's use the first equation:
5 + y = 4 y = 4 - 5 y = -1
So, y = -1.
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Check the solution: Substitute x = 5 and y = -1 into both original equations to verify the solution:
- Equation 1: 5 + (-1) = 4 (Correct)
- Equation 2: 5 - (-1) = 6 (Correct)
The solution satisfies both equations.
Therefore, the solution to the system of equations is (5, -1). This means that the point (5, -1) is the intersection of the two lines represented by the equations x + y = 4 and x - y = 6 on a coordinate plane. The elimination method provides a systematic way to arrive at this solution, ensuring accuracy and efficiency.
Conclusion
Solving systems of linear equations is a fundamental skill in mathematics with wide-ranging applications. The elimination method offers a powerful and efficient approach to finding solutions, particularly when dealing with equations where the coefficients of one variable are opposites or can be easily made opposites. By following the steps outlined in this guide – aligning equations, making coefficients opposites, adding equations, solving for the remaining variable, substituting back, and checking the solution – you can confidently solve a wide variety of systems of linear equations. Understanding and mastering this method provides a solid foundation for more advanced mathematical concepts and problem-solving in various fields. The ability to solve these systems is not just an academic exercise; it is a crucial tool for modeling and analyzing real-world scenarios in science, engineering, economics, and beyond.
Answer
The correct answer is C. (5, -1).