Solving Systems Of Equations Using Elimination Method

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In the realm of mathematics, solving systems of equations is a fundamental skill with widespread applications in various fields, including engineering, economics, and computer science. A system of equations is a set of two or more equations that share the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. In this comprehensive guide, we will delve into the intricacies of solving a specific system of equations using the elimination method, providing a step-by-step approach to ensure clarity and understanding. Mastering the elimination method is crucial for efficiently tackling complex mathematical problems and gaining a deeper appreciation for the elegance of algebraic solutions.

The specific system of equations we will address is:

$\left\{\begin{array}{l}-5 x+7 y=31 \\ 3 x+4 y=6\end{array}\right.$

This system consists of two linear equations with two variables, x and y. Our goal is to find the values of x and y that satisfy both equations concurrently. To achieve this, we will employ the elimination method, a powerful technique that involves manipulating the equations to eliminate one variable, allowing us to solve for the other.

The Elimination Method: A Detailed Walkthrough

The elimination method, also known as the addition method, is a technique used to solve systems of linear equations by eliminating one of the variables. This is achieved by multiplying one or both equations by constants so that the coefficients of one variable are opposites. When the equations are added together, this variable is eliminated, leaving an equation in only one variable that can be easily solved. This method is particularly effective when the coefficients of one variable are easily made opposites or when dealing with larger systems of equations.

The elimination method hinges on the principle that multiplying an equation by a non-zero constant does not change its solution set. Similarly, adding two equations together results in a new equation that shares the same solutions as the original system. By strategically applying these operations, we can transform the system into a more manageable form.

Step 1: Multiplying Equations to Match Coefficients

The first step in the elimination method is to manipulate the equations so that the coefficients of one variable are opposites or multiples of each other. This allows for the elimination of that variable when the equations are added. In our system, we can choose to eliminate either x or y. Let's choose to eliminate x. To do this, we need to find the least common multiple (LCM) of the coefficients of x, which are -5 and 3. The LCM of 5 and 3 is 15. Therefore, we want to multiply the equations by constants that will make the coefficients of x equal to 15 and -15.

To make the coefficient of x in the first equation -15, we multiply the entire equation by 3:

3 * (-5x + 7y) = 3 * 31

This simplifies to:

-15x + 21y = 93

Similarly, to make the coefficient of x in the second equation 15, we multiply the entire equation by 5:

5 * (3x + 4y) = 5 * 6

This simplifies to:

15x + 20y = 30

Now our system of equations looks like this:

$\left\{\begin{array}{l}-15 x+21 y=93 \\ 15 x+20 y=30\end{array}\right.$

Notice that the coefficients of x are now opposites (-15 and 15). This sets the stage for the next step, where we will add the equations together to eliminate x.

Step 2: Eliminating a Variable by Adding Equations

With the coefficients of x now opposites, we can proceed to eliminate x by adding the two equations together. When we add the left-hand sides of the equations, the -15x and 15x terms cancel each other out, leaving us with an equation in terms of y only. Adding the right-hand sides gives us a constant value. This process effectively reduces the system of two equations with two variables into a single equation with one variable, making it much easier to solve.

Adding the equations:

(-15x + 21y) + (15x + 20y) = 93 + 30

Simplifying, we get:

41y = 123

Now we have a simple equation with only one variable, y. This is a significant step forward in solving the system, as we can now isolate y and find its value.

Step 3: Solving for the Remaining Variable

After eliminating one variable, we are left with a single equation in one variable. In our case, we have the equation 41y = 123. To solve for y, we simply divide both sides of the equation by the coefficient of y, which is 41. This isolates y and gives us its value.

Dividing both sides by 41:

y = 123 / 41

This gives us:

y = 3

We have now found the value of y, which is 3. This is a crucial piece of the solution to the system of equations. We can use this value to find the value of x by substituting it back into one of the original equations.

Step 4: Substituting to Find the Other Variable

Now that we have the value of y, we can substitute it back into either of the original equations to solve for x. It's generally a good practice to choose the equation that looks simpler or easier to work with. Let's substitute y = 3 into the second original equation, 3x + 4y = 6.

Substituting y = 3:

3x + 4(3) = 6

Simplifying, we get:

3x + 12 = 6

To isolate x, we first subtract 12 from both sides of the equation:

3x = 6 - 12

3x = -6

Now, we divide both sides by 3:

x = -6 / 3

This gives us:

x = -2

We have now found the value of x, which is -2. This completes the solution to the system of equations.

Step 5: Verifying the Solution

To ensure the accuracy of our solution, it is essential to verify that the values of x and y satisfy both original equations. This step helps to catch any potential errors made during the solving process. We substitute the values x = -2 and y = 3 into both equations and check if the equations hold true.

Let's verify the first equation, -5x + 7y = 31:

-5(-2) + 7(3) = 31

10 + 21 = 31

31 = 31

The first equation is satisfied.

Now, let's verify the second equation, 3x + 4y = 6:

3(-2) + 4(3) = 6

-6 + 12 = 6

6 = 6

The second equation is also satisfied.

Since both equations are satisfied by x = -2 and y = 3, we can confidently conclude that this is the correct solution to the system of equations. This verification step underscores the importance of thoroughness in mathematical problem-solving.

The Solution and its Interpretation

Having completed the elimination method, we have arrived at the solution to the system of equations:

$\left\{\begin{array}{l}x = -2 \\ y = 3\end{array}\right.$

This means that the point (-2, 3) is the intersection of the two lines represented by the equations -5x + 7y = 31 and 3x + 4y = 6. In graphical terms, this solution represents the coordinates of the point where the two lines intersect on the Cartesian plane. This intersection point is the unique solution that satisfies both equations simultaneously.

In summary, the solution to the system of equations is x = -2 and y = 3.

Inconsistent and Dependent Systems

While the system we solved had a unique solution, it's important to understand that not all systems of equations behave this way. There are two other possibilities:

• Inconsistent Systems: An inconsistent system is a system of equations that has no solution. This occurs when the lines represented by the equations are parallel and never intersect. In the elimination method, an inconsistent system will result in a contradiction, such as 0 = a non-zero number. For example, if during the elimination process, you arrive at an equation like 0 = 5, this indicates that the system is inconsistent.
• Dependent Systems: A dependent system is a system of equations that has infinitely many solutions. This occurs when the equations represent the same line. In the elimination method, a dependent system will result in an identity, such as 0 = 0. This means that the equations are essentially multiples of each other, and any solution to one equation will also be a solution to the other. For instance, if you end up with an equation like 0 = 0 after elimination, the system is dependent.

In our case, since we found a unique solution, the system is neither inconsistent nor dependent. It is a consistent and independent system.

Conclusion: Mastering the Elimination Method

In this comprehensive guide, we have meticulously explored the process of solving a system of linear equations using the elimination method. We have broken down each step, from manipulating equations to eliminate a variable to verifying the solution. By following this step-by-step approach, you can confidently tackle similar problems and develop a strong foundation in solving systems of equations.

The elimination method is a versatile and powerful tool in the realm of mathematics. It not only provides a systematic way to solve systems of equations but also enhances your understanding of algebraic manipulations and problem-solving strategies. By mastering this method, you will be well-equipped to tackle a wide range of mathematical challenges and appreciate the beauty and elegance of algebraic solutions. Remember to practice consistently, and you will undoubtedly become proficient in solving systems of equations using the elimination method.

Furthermore, understanding the concepts of inconsistent and dependent systems is crucial for a complete understanding of systems of equations. Recognizing these cases allows you to interpret the nature of the solutions and avoid unnecessary calculations. By grasping these nuances, you can approach systems of equations with greater confidence and accuracy. The elimination method is not just a technique; it's a pathway to deeper mathematical understanding and problem-solving prowess.