Solving Systems Of Equations A Detailed Guide
Introduction
In the realm of mathematics, solving systems of equations is a fundamental skill with wide-ranging applications across various fields, including engineering, economics, and computer science. A system of equations is a collection of two or more equations with the same set of variables. The goal is to find values for the variables that satisfy all equations simultaneously. In this article, we will delve into solving a system of two linear equations using different methods, providing a comprehensive understanding of the process. Specifically, we will tackle the following system:
y = -0.95x + 0.644
y = -0.21x + 0.5552
This system consists of two linear equations, each representing a straight line on a coordinate plane. The solution to the system is the point (or points) where these lines intersect. There are several methods to solve such systems, including substitution, elimination, and graphical methods. We will primarily focus on the substitution and elimination methods in this discussion, as they offer precise algebraic solutions. Understanding these methods is crucial for anyone studying algebra or related fields.
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be easily solved. Once we find the value of one variable, we can substitute it back into either of the original equations to find the value of the other variable. This method is particularly useful when one of the equations is already solved for one variable, as is the case in our system. The efficiency and directness of the substitution method make it a valuable tool in solving systems of equations.
On the other hand, the elimination method, also known as the addition method, involves manipulating the equations so that the coefficients of one variable are opposites. When the equations are added, that variable is eliminated, resulting in a single equation with one variable. This method is effective when the coefficients of one variable are easily made opposites by multiplying one or both equations by a constant. After solving for one variable, we can substitute the value back into one of the original equations to find the other variable. The elimination method is advantageous for its systematic approach and is particularly useful when dealing with more complex systems of equations.
Solving the System
To effectively solve the system of equations, we can employ several methods, each offering a unique approach to finding the solution. The system we are addressing is:
y = -0.95x + 0.644
y = -0.21x + 0.5552
This system presents an opportunity to use both the substitution and elimination methods. Let's start by applying the substitution method, which is particularly straightforward in this case since both equations are already solved for y. By setting the two expressions for y equal to each other, we can eliminate y and solve for x.
Substitution Method
The substitution method involves solving one equation for one variable and substituting that expression into the other equation. Since both equations are already solved for y, we can set them equal to each other:
-0.95x + 0.644 = -0.21x + 0.5552
This equation now contains only one variable, x, which we can solve for. To do this, we first want to isolate the x terms on one side of the equation and the constants on the other side. We can add 0.95x to both sides and subtract 0.5552 from both sides:
-0.95x + 0.644 + 0.95x - 0.5552 = -0.21x + 0.5552 + 0.95x - 0.5552
Simplifying this gives us:
0. 0888 = 0.74x
Now, to solve for x, we divide both sides by 0.74:
x = 0.0888 / 0.74
x = 0.12
So, we have found that x = 0.12. Next, we substitute this value back into one of the original equations to solve for y. Let's use the first equation:
y = -0.95x + 0.644
y = -0.95(0.12) + 0.644
y = -0.114 + 0.644
y = 0.53
Thus, we find that y = 0.53. Therefore, the solution to the system of equations is the point (0.12, 0.53). The substitution method allows us to systematically reduce the system to a single equation, making it easier to solve. This approach highlights the power of algebraic manipulation in finding solutions to complex problems.
Elimination Method
The elimination method, also known as the addition method, provides an alternative approach to solving systems of equations. While the substitution method is quite straightforward in this case, demonstrating the elimination method offers a broader understanding of equation-solving techniques. To apply the elimination method, we need to manipulate the equations so that the coefficients of one variable are opposites. When we add the equations, that variable will be eliminated, leaving us with a single equation in one variable.
Let's revisit our system of equations:
y = -0.95x + 0.644
y = -0.21x + 0.5552
To use the elimination method effectively, we first rewrite the equations in the standard form Ax + By = C. This involves moving the x terms to the left side of the equations:
0. 95x + y = 0.644
1. 21x + y = 0.5552
Now, we want to eliminate one of the variables. Notice that the coefficients of y are both 1. To eliminate y, we can multiply one of the equations by -1 and then add the equations together. Let's multiply the second equation by -1:
-1(0.21x + y) = -1(0.5552)
-0.21x - y = -0.5552
Now, we add the first equation to the modified second equation:
(0.95x + y) + (-0.21x - y) = 0.644 + (-0.5552)
Simplifying this, we get:
2. 74x = 0.0888
Dividing both sides by 0.74, we find:
x = 0.0888 / 0.74
x = 0.12
This matches the x value we found using the substitution method. Now, we substitute x = 0.12 back into one of the original equations to solve for y. Let's use the first original equation:
y = -0.95x + 0.644
y = -0.95(0.12) + 0.644
y = -0.114 + 0.644
y = 0.53
Again, we find that y = 0.53. Thus, the solution to the system of equations, using the elimination method, is also the point (0.12, 0.53). The elimination method provides a structured way to solve systems by systematically eliminating variables. This method is particularly useful when dealing with larger systems of equations where substitution might become cumbersome.
Conclusion
In conclusion, we have successfully solved the given system of equations using both the substitution and elimination methods. The system we addressed was:
y = -0.95x + 0.644
y = -0.21x + 0.5552
Through both methods, we found that the solution to this system is the point (0.12, 0.53). This means that the lines represented by these equations intersect at this point on the coordinate plane. The fact that both methods yielded the same solution reinforces the accuracy and validity of our results. Understanding and applying these methods is crucial for solving systems of equations in various mathematical and real-world contexts.
The substitution method involved setting the two expressions for y equal to each other, solving for x, and then substituting the value of x back into one of the equations to find y. This method is particularly efficient when one or both equations are already solved for one variable, as was the case here. The step-by-step approach ensures that each variable is accurately determined, leading to a precise solution.
On the other hand, the elimination method required us to rewrite the equations in standard form, manipulate them to have opposite coefficients for one variable, and then add the equations to eliminate that variable. After solving for the remaining variable, we substituted its value back into one of the original equations to find the other variable. The elimination method is particularly useful when dealing with systems where the coefficients are easily manipulated to create opposites. This method provides a systematic way to reduce the system to a single equation, simplifying the solution process.
The importance of solving systems of equations extends beyond the classroom. In real-world applications, systems of equations are used to model and solve problems in various fields, including economics, engineering, physics, and computer science. For example, economists use systems of equations to model supply and demand, engineers use them to design structures and circuits, and physicists use them to describe the motion of objects. Mastering the techniques to solve these systems equips individuals with powerful tools for problem-solving and decision-making.
In summary, both the substitution and elimination methods are valuable tools for solving systems of equations. The choice of method often depends on the specific structure of the equations, but both methods provide accurate solutions when applied correctly. The solution to the system:
y = -0.95x + 0.644
y = -0.21x + 0.5552
is the point (0.12, 0.53), which represents the intersection of the two lines. This exercise underscores the significance of understanding and applying different algebraic techniques to solve mathematical problems efficiently and accurately.
