Solving Systems Of Equations By Elimination A Step-by-Step Guide

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In the realm of mathematics, solving systems of equations is a fundamental skill. One effective method for tackling these systems is elimination, where the goal is to manipulate the equations in such a way that one variable is eliminated, simplifying the problem. This article delves into how Andre might approach solving a specific system of equations using this method, offering insights into the underlying principles and potential outcomes.

The given system of equations is:

{3a+6b=12−3a+6b=−12\left\{ \begin{array}{c} 3a + 6b = 12 \\ -3a + 6b = -12 \end{array} \right.

Understanding Elimination

The elimination method hinges on the idea of adding or subtracting multiples of equations to cancel out one of the variables. This is possible when the coefficients of one variable in the two equations are either the same or additive inverses (opposites). By strategically combining the equations, we can create a new equation with only one variable, which can then be easily solved. The solution for this variable can then be substituted back into one of the original equations to find the value of the other variable.

In Andre's case, we observe that the coefficients of 'a' in the two equations are 3 and -3, which are additive inverses. This makes the elimination method particularly straightforward in this scenario. Let's explore the steps Andre might take.

Step-by-Step Elimination

Andre's first step would likely be to add the two equations together. This is because the 'a' terms will cancel out:

(3a+6b)+(−3a+6b)=12+(−12)(3a + 6b) + (-3a + 6b) = 12 + (-12)

Combining like terms, we get:

3a−3a+6b+6b=12−123a - 3a + 6b + 6b = 12 - 12

0a+12b=00a + 12b = 0

12b=012b = 0

This resulting equation, 12b = 0, is a crucial step in solving the system. It immediately tells us that the value of 'b' must be 0. This is because any number multiplied by 12 to give 0 must itself be 0. Therefore, Andre has successfully eliminated the variable 'a' and arrived at a simple equation for 'b'.

Solving for b

To find the value of 'b', Andre would simply divide both sides of the equation by 12:

12b12=012\frac{12b}{12} = \frac{0}{12}

b=0b = 0

So, Andre has determined that b = 0. Now, he needs to find the value of 'a'.

Solving for a

To find 'a', Andre can substitute the value of 'b' (which is 0) into either of the original equations. Let's use the first equation:

3a+6b=123a + 6b = 12

Substitute b = 0:

3a+6(0)=123a + 6(0) = 12

3a+0=123a + 0 = 12

3a=123a = 12

Now, divide both sides by 3:

3a3=123\frac{3a}{3} = \frac{12}{3}

a=4a = 4

Thus, Andre finds that a = 4. Therefore, the solution to the system of equations is a = 4 and b = 0.

Verification

It's always a good practice to verify the solution by substituting the values of 'a' and 'b' back into both original equations:

For the first equation:

3a+6b=123a + 6b = 12

3(4)+6(0)=123(4) + 6(0) = 12

12+0=1212 + 0 = 12

12=1212 = 12

This holds true.

For the second equation:

−3a+6b=−12-3a + 6b = -12

−3(4)+6(0)=−12-3(4) + 6(0) = -12

−12+0=−12-12 + 0 = -12

−12=−12-12 = -12

This also holds true. The solution a = 4 and b = 0 satisfies both equations, confirming its correctness.

Possible Resulting Equations and Implications

As we've seen, one possible resulting equation when Andre eliminates a variable is 12b = 0. This equation directly leads to the solution for 'b'. However, it's important to understand the implications of different resulting equations. For instance:

  • If the resulting equation were of the form 0 = 0, this would indicate that the two original equations are dependent, meaning they represent the same line. In this case, there would be infinitely many solutions.
  • If the resulting equation were of the form 0 = a non-zero number, this would indicate that the system of equations has no solution because it represents a contradiction.

In Andre's specific case, the resulting equation 12b = 0 is straightforward and leads to a unique solution. This highlights the power of the elimination method in simplifying systems of equations and revealing the relationships between variables.

Alternative Approaches and Considerations

While elimination is effective here, it's worth noting that other methods, such as substitution, could also be used to solve this system. The choice of method often depends on the specific structure of the equations. In this case, elimination is particularly efficient because the 'a' terms have opposite coefficients.

Furthermore, it's important to consider the potential for errors when solving systems of equations. Careful attention to detail and methodical steps are crucial to avoid mistakes in arithmetic and algebraic manipulation. Verification, as demonstrated earlier, is a valuable tool for ensuring the accuracy of the solution.

The Broader Context of Systems of Equations

Systems of equations are not just abstract mathematical concepts; they have wide-ranging applications in various fields. They are used to model and solve problems in physics, engineering, economics, computer science, and many other disciplines. Understanding how to solve these systems is therefore an essential skill for anyone pursuing studies or careers in these areas.

For example, in physics, systems of equations can be used to analyze the motion of objects under multiple forces. In economics, they can be used to model supply and demand in markets. In computer science, they can be used in algorithms for optimization and machine learning. The ability to solve systems of equations provides a powerful toolkit for tackling real-world problems.

The Importance of Understanding Methods

While tools and calculators can assist in solving systems of equations, a deep understanding of the underlying methods is crucial. This understanding allows one to choose the most appropriate method for a given problem, to interpret the results correctly, and to identify potential errors. Moreover, it fosters a broader mathematical literacy that is valuable in many contexts.

The elimination method, as demonstrated by Andre's approach, is a prime example of such a fundamental method. It showcases the power of algebraic manipulation and the elegance of mathematical reasoning. By mastering this method and others, individuals can become more confident and effective problem-solvers.

Conclusion

Andre's approach to solving the system of equations using elimination illustrates a powerful and versatile technique in algebra. By strategically adding the equations, he eliminated one variable and arrived at a simple equation that readily yielded the value of 'b'. Subsequent substitution allowed him to find the value of 'a', completing the solution. This process underscores the importance of understanding the principles behind elimination and the careful steps required to arrive at a correct solution. The resulting equation 12b = 0 is a direct consequence of the elimination process and highlights the effectiveness of this method in solving systems of equations. This exploration not only provides a solution to the specific problem but also offers a broader understanding of the role and application of systems of equations in various fields.

The core question presented is: Which could be the resulting equations when André eliminates one of the variables in the system?

{3a+6b=12−3a+6b=−12\left\{ \begin{array}{c} 3 a+6 b=12 \\ -3 a+6 b=-12 \end{array} \right.

We've already determined that by adding the two equations, Andre would obtain 12b = 0. So, let's analyze this result and other possibilities.

Detailed Analysis of Elimination Steps

To reiterate, Andre's goal is to eliminate one of the variables, either 'a' or 'b', to simplify the system of equations. The provided system has a structure that lends itself well to elimination due to the coefficients of 'a' being additive inverses (3 and -3). Let's break down the process step by step:

  1. Adding the Equations:

    • This is the most straightforward approach given the structure of the equations.
    • When the two equations are added, the 'a' terms cancel each other out:

      (3a+6b)+(−3a+6b)=12+(−12)(3a + 6b) + (-3a + 6b) = 12 + (-12)

      3a−3a+6b+6b=12−123a - 3a + 6b + 6b = 12 - 12

      12b=012b = 0

    • This resulting equation, 12b = 0, is indeed a valid outcome of the elimination process.

  2. Alternative Elimination (Subtracting Equations):

    • Although adding the equations is the most direct approach, let's consider what happens if Andre were to subtract the second equation from the first.
    • Subtracting the equations:

      (3a+6b)−(−3a+6b)=12−(−12)(3a + 6b) - (-3a + 6b) = 12 - (-12)

      3a+6b+3a−6b=12+123a + 6b + 3a - 6b = 12 + 12

      6a=246a = 24

    • This results in another valid equation where the 'b' variable has been eliminated: 6a = 24.

Analyzing the Given Options

Now, let's consider the options provided in the original problem:

A. 12b = 0

  • As we've shown, this is a possible resulting equation when Andre eliminates 'a' by adding the two equations together. Thus, this is a correct option.

Deeper Dive into the Implications

The resulting equations from elimination are not just intermediate steps; they provide valuable insights into the nature of the system of equations. In this case, both 12b = 0 and 6a = 24 offer direct pathways to solving for the variables.

  • 12b = 0 immediately implies that b = 0. This is because the only number that, when multiplied by 12, results in 0 is 0 itself.
  • 6a = 24 similarly implies that a = 4. Dividing both sides of the equation by 6 gives a = 24/6 = 4.

These resulting equations are crucial because they reduce a two-variable problem into two single-variable problems, making the solution process significantly easier.

The Significance of Choosing the Right Operation

Andre's success in eliminating a variable hinges on choosing the correct operation (addition or subtraction) based on the structure of the equations. In this case, adding the equations was the most intuitive choice because the coefficients of 'a' were already additive inverses. However, if the coefficients were not additive inverses or the same, Andre might need to multiply one or both equations by a constant before adding or subtracting them. This is a key aspect of the elimination method – manipulating the equations to create opportunities for cancellation.

The Role of Verification

Once Andre obtains a solution (values for 'a' and 'b'), it's essential to verify that the solution satisfies both original equations. This step helps to catch any errors made during the elimination or solving process. As we demonstrated earlier, substituting a = 4 and b = 0 into the original equations confirms that these values are indeed the correct solution.

Connecting Elimination to Other Methods

The elimination method is one of several techniques for solving systems of equations. Other common methods include substitution and graphing. The choice of method often depends on the specific problem. Elimination is particularly effective when the coefficients of one variable are easily made additive inverses or the same. Substitution is useful when one equation can be easily solved for one variable in terms of the other. Graphing provides a visual representation of the equations and their intersection points, which represent the solutions.

Understanding multiple methods for solving systems of equations is crucial for developing mathematical flexibility and problem-solving skills.

Real-World Applications and Context

Systems of equations are fundamental to modeling and solving problems in various real-world contexts. They appear in fields such as physics, engineering, economics, computer science, and statistics. For example, in physics, systems of equations can describe the motion of objects under multiple forces. In economics, they can model supply and demand curves. In computer science, they are used in optimization algorithms and machine learning.

Therefore, mastering the techniques for solving systems of equations, including elimination, is an essential skill for anyone pursuing studies or careers in these areas.

The Broader Mathematical Landscape

The elimination method is not just a standalone technique; it's part of a broader mathematical landscape that includes linear algebra, matrix operations, and other advanced concepts. In more complex systems of equations, especially those with many variables, techniques from linear algebra, such as Gaussian elimination and matrix inversion, become essential. Understanding the fundamental principles of elimination provides a solid foundation for these more advanced topics.

Refining the Understanding of Elimination

The process of elimination is rooted in the fundamental properties of equality. Specifically, it relies on the fact that adding the same quantity to both sides of an equation preserves the equality. Similarly, multiplying both sides of an equation by a non-zero constant also preserves the equality. These principles allow us to manipulate equations strategically to eliminate variables.

Conclusion: Identifying the Resulting Equation

In summary, when Andre eliminates one of the variables in the given system of equations, 12b = 0 is a valid and likely resulting equation. This equation is obtained by adding the two original equations, which eliminates the 'a' variable. This highlights the effectiveness of the elimination method in simplifying systems of equations and providing a direct path to the solution.