Solving System Of Equations 5x + 2y - 3 = 0 And X + 2y = 5

This article provides a detailed explanation of how to solve a system of two linear equations. Specifically, we will focus on the equations 5x + 2y - 3 = 0 and x + 2y = 5. We will explore various methods, including substitution and elimination, to find the values of x and y that satisfy both equations simultaneously. Understanding how to solve systems of equations is a fundamental skill in mathematics with applications in various fields such as physics, engineering, and economics. By the end of this article, you will have a clear understanding of the steps involved and be able to apply these techniques to solve similar problems.

Introduction to Systems of Linear Equations

A system of linear equations is a set of two or more linear equations containing the same variables. The solution to a system of linear equations is the set of values for the variables that make all the equations true. Graphically, the solution represents the point(s) where the lines corresponding to the equations intersect. If the lines are parallel, there is no solution, and if the lines are the same, there are infinitely many solutions. In this case, we are dealing with two equations in two variables, x and y. Our goal is to find the unique pair of values (x, y) that satisfies both 5x + 2y - 3 = 0 and x + 2y = 5. Solving systems of equations is a cornerstone of algebra and has far-reaching applications in modeling real-world problems. For instance, in economics, these systems can represent supply and demand curves, and the solution can represent the market equilibrium point. In engineering, they can be used to analyze circuits or structural systems. Therefore, mastering the techniques for solving these systems is crucial for anyone pursuing studies or careers in STEM fields.

Method 1: Solving by Substitution

One effective method for solving systems of equations is the substitution method. This method involves solving one equation for one variable and then substituting that expression into the other equation. Let's apply this to our system: 5x + 2y - 3 = 0 and x + 2y = 5. First, we can solve the second equation for x:

x + 2y = 5 x = 5 - 2y

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

5(5 - 2y) + 2y - 3 = 0 25 - 10y + 2y - 3 = 0 22 - 8y = 0 8y = 22 y = 22/8 = 11/4

Now that we have the value of y, we can substitute it back into the equation x = 5 - 2y to find the value of x:

x = 5 - 2(11/4) x = 5 - 11/2 x = 10/2 - 11/2 x = -1/2

Therefore, the solution to the system of equations using the substitution method is x = -1/2 and y = 11/4. The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. This approach simplifies the problem by reducing it to a single equation in one variable, which can then be solved using standard algebraic techniques. The key to success with the substitution method is to choose the equation and variable that will lead to the simplest expression for substitution. In more complex systems, this can significantly reduce the amount of algebraic manipulation required.

Method 2: Solving by Elimination

The elimination method, also known as the addition method, is another powerful technique for solving systems of equations. This method involves manipulating the equations so that the coefficients of one of the variables are opposites, and then adding the equations together to eliminate that variable. Let’s apply this method to our system: 5x + 2y - 3 = 0 and x + 2y = 5. First, we rewrite the first equation as 5x + 2y = 3. Now we have:

5x + 2y = 3 x + 2y = 5

To eliminate y, we can multiply the second equation by -1:

5x + 2y = 3 -x - 2y = -5

Now, we add the two equations together:

(5x + 2y) + (-x - 2y) = 3 + (-5) 4x = -2 x = -2/4 = -1/2

Now that we have the value of x, we can substitute it back into either of the original equations to find the value of y. Let's use the second equation, x + 2y = 5:

(-1/2) + 2y = 5 2y = 5 + 1/2 2y = 11/2 y = (11/2) / 2 y = 11/4

Therefore, the solution to the system of equations using the elimination method is x = -1/2 and y = 11/4. The elimination method is particularly advantageous when the coefficients of one of the variables are already close to being opposites or can be easily made so by multiplying one or both equations by a constant. This method is often preferred when dealing with larger systems of equations, as it can be more efficient than substitution in such cases. The core idea behind the elimination method is to strategically manipulate the equations to create a situation where adding them together results in the cancellation of one variable, thereby simplifying the problem.

Verification of the Solution

After solving a system of equations, it's crucial to verify the solution to ensure its correctness. This involves substituting the values of x and y back into both original equations to check if they hold true. Our solution is x = -1/2 and y = 11/4. Let's substitute these values into the first equation, 5x + 2y - 3 = 0:

5(-1/2) + 2(11/4) - 3 = 0 -5/2 + 11/2 - 3 = 0 6/2 - 3 = 0 3 - 3 = 0 0 = 0

The first equation holds true. Now, let's substitute the values into the second equation, x + 2y = 5:

(-1/2) + 2(11/4) = 5 -1/2 + 11/2 = 5 10/2 = 5 5 = 5

The second equation also holds true. Since the values x = -1/2 and y = 11/4 satisfy both equations, we can confidently conclude that this is the correct solution to the system. Verifying the solution is an essential step in the problem-solving process, as it helps to catch any errors that might have occurred during the algebraic manipulations. This practice not only ensures the accuracy of the solution but also reinforces understanding of the underlying concepts and techniques involved in solving systems of equations. In more complex scenarios, where solutions may not be easily apparent, verification becomes even more critical.

Graphical Interpretation

Visualizing the solution to a system of equations graphically can provide a deeper understanding of the problem. Each linear equation represents a line on the coordinate plane. The solution to the system is the point where the two lines intersect. Let's consider our equations: 5x + 2y - 3 = 0 and x + 2y = 5. We found the solution to be x = -1/2 and y = 11/4, which corresponds to the point (-0.5, 2.75) on the coordinate plane. To graph the equations, we can rewrite them in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

For the first equation, 5x + 2y - 3 = 0, we can rewrite it as:

2y = -5x + 3 y = (-5/2)x + 3/2

This line has a slope of -5/2 and a y-intercept of 3/2 (1.5).

For the second equation, x + 2y = 5, we can rewrite it as:

2y = -x + 5 y = (-1/2)x + 5/2

This line has a slope of -1/2 and a y-intercept of 5/2 (2.5).

If you were to plot these two lines on a graph, you would see that they intersect at the point (-0.5, 2.75), confirming our algebraic solution. The graphical interpretation provides a visual confirmation of the solution and highlights the relationship between the algebraic and geometric representations of linear equations. This approach is particularly useful for understanding systems with no solution (parallel lines) or infinitely many solutions (coinciding lines). Furthermore, graphing systems of equations can be a valuable tool in applied mathematics, where solutions often represent real-world quantities and their relationships.

Conclusion

In this article, we have thoroughly discussed how to solve the system of equations 5x + 2y - 3 = 0 and x + 2y = 5. We explored two primary methods: substitution and elimination, both of which led us to the same solution: x = -1/2 and y = 11/4. We then verified this solution by substituting the values back into the original equations, confirming its accuracy. Additionally, we examined the graphical interpretation of the solution, which provided a visual confirmation of the intersection point of the two lines represented by the equations. Mastering these techniques for solving systems of linear equations is crucial for various mathematical and real-world applications. The ability to solve these systems efficiently and accurately is a valuable skill in fields such as engineering, physics, economics, and computer science. Whether you prefer the substitution method, the elimination method, or a graphical approach, understanding the underlying principles and practicing these techniques will enhance your problem-solving abilities. Furthermore, the concepts discussed here serve as a foundation for more advanced topics in linear algebra and calculus, making it essential for continued mathematical study and application. The knowledge gained from solving systems of equations extends beyond the classroom, enabling you to model and analyze various scenarios in everyday life and professional settings.