Solving System Of Equations By Elimination Finding The Value Of X

In mathematics, systems of linear equations are a fundamental concept, especially in algebra. These systems consist of two or more linear equations that involve the same set of variables. Solving such systems means finding the values of the variables that satisfy all equations simultaneously. This is a crucial skill with applications across various fields, from engineering to economics. There are several methods to solve these systems, including substitution, elimination, and graphical methods. Each method has its strengths, and the choice often depends on the specific equations at hand. The elimination method, which Veda used in this problem, is particularly effective when the coefficients of one variable in the equations are multiples of each other or when they have opposite signs, as it allows for the easy elimination of one variable, simplifying the system and making it easier to solve for the remaining variable. This method involves manipulating the equations (by multiplying them by constants) so that, when added or subtracted, one variable cancels out, leaving a single equation with one variable. This resulting equation can then be easily solved, and the solution can be substituted back into the original equations to find the values of the other variables.

Understanding the underlying principles of these methods is crucial for success in algebra and beyond. The solution to a system of linear equations represents the point (or points) where the lines (or planes, in higher dimensions) represented by the equations intersect. This geometric interpretation provides a valuable visual aid in understanding the algebraic manipulations involved in solving the system. Moreover, the ability to solve systems of linear equations is a prerequisite for more advanced topics in mathematics, such as linear algebra, calculus, and differential equations. Proficiency in this area is not only beneficial for academic pursuits but also for various real-world applications where mathematical models are used to solve problems.

The specific problem at hand involves a system of two linear equations, and our goal is to find the value of x in the solution. The system is given as:

$$\left\{ \begin{aligned} 6 + 4x - 2y &= 0 \\ -3 - 7y &= 10x \end{aligned} \right.$$

Veda has used the method of elimination to solve this system, and we are tasked with determining the numerical value of x that satisfies both equations. This requires a systematic approach to manipulating the equations and isolating x. The process will involve rearranging the equations into a more standard form, then strategically multiplying one or both equations by constants to make the coefficients of either x or y the same (or additive inverses). This will allow us to eliminate one variable by adding or subtracting the equations, thereby reducing the system to a single equation in one variable. Solving this single equation will give us the value of one variable, which can then be substituted back into either of the original equations to find the value of the other variable.

This type of problem is a classic example of how linear algebra principles are applied in practice. The ability to solve such systems is a key skill in many fields, including engineering, physics, economics, and computer science. The challenge here is not just to find the solution but also to understand the process and the underlying mathematical concepts. By carefully following the steps of the elimination method and paying attention to algebraic details, we can arrive at the correct solution for x. The importance of this skill extends beyond the classroom, as linear equations and their solutions are fundamental tools in modeling and solving real-world problems.

To solve the system of equations using the elimination method, we first need to rearrange the equations into a more standard form, aligning the variables and constants. The given system is:

$$\left\{ \begin{aligned} 6 + 4x - 2y &= 0 \\ -3 - 7y &= 10x \end{aligned} \right.$$

Rearranging the first equation, we get:

$$4x - 2y = -6$$

Rearranging the second equation, we get:

$$-10x - 7y = 3$$

Now our system looks like this:

$$\left\{ \begin{aligned} 4x - 2y &= -6 \\ -10x - 7y &= 3 \end{aligned} \right.$$

The next step is to eliminate one of the variables. Let's eliminate x. To do this, we need to make the coefficients of x in both equations additive inverses. We can multiply the first equation by 5 and the second equation by 2:

Multiplying the first equation by 5:

$$5(4x - 2y) = 5(-6) \Rightarrow 20x - 10y = -30$$

Multiplying the second equation by 2:

$$2(-10x - 7y) = 2(3) \Rightarrow -20x - 14y = 6$$

Now our system is:

$$\left\{ \begin{aligned} 20x - 10y &= -30 \\ -20x - 14y &= 6 \end{aligned} \right.$$

Adding the two equations will eliminate x:

$$(20x - 10y) + (-20x - 14y) = -30 + 6$$

$$-24y = -24$$

Now, solve for y:

$$y = \frac{-24}{-24} = 1$$

Now that we have the value of y, we can substitute it back into one of the original equations to find the value of x. Let's use the first rearranged equation:

$$4x - 2y = -6$$

Substitute y = 1:

$$4x - 2(1) = -6$$

$$4x - 2 = -6$$

Add 2 to both sides:

$$4x = -4$$

Divide by 4:

$$x = -1$$

Thus, the value of x in the solution of the system of equations is -1. This methodical approach, breaking down the problem into smaller, manageable steps, is crucial for solving complex mathematical problems. The ability to manipulate equations, identify opportunities for elimination, and accurately perform algebraic operations is a cornerstone of mathematical proficiency. The solution not only provides the answer but also reinforces the importance of systematic problem-solving skills.

After meticulously applying the elimination method to the given system of equations, we have successfully determined the value of x. By rearranging the equations, strategically multiplying them to align the coefficients of x, and then eliminating x through addition, we were able to solve for y. Subsequently, substituting the value of y back into one of the original equations allowed us to isolate and solve for x. The detailed steps taken in the solution process demonstrate the power and precision of algebraic manipulation in solving systems of linear equations.

The final answer, obtained through this systematic approach, is:

$$x = -1$$

This result confirms that x equals -1 in the solution of the given system of equations. This process highlights the importance of a clear understanding of algebraic principles and the ability to apply them methodically. The solution not only provides a numerical answer but also reinforces the broader skills of problem-solving, logical reasoning, and attention to detail. These skills are essential not only in mathematics but also in various other disciplines and real-world applications. The ability to solve systems of equations is a fundamental tool for anyone working in quantitative fields, and mastering the elimination method is a key step in developing this skill.

The elimination method is a powerful technique for solving systems of linear equations. It's particularly useful when the equations have coefficients that are easily made additive inverses of each other. The core idea behind this method is to manipulate the equations in such a way that, when they are added together, one of the variables cancels out, leaving a single equation with only one variable. This resulting equation can then be easily solved, and the solution can be substituted back into the original equations to find the values of the other variables. This process not only simplifies the system but also provides a clear path to finding the solution.

The first step in the elimination method is to align the equations in a standard format, with like terms (i.e., the x terms, y terms, and constants) lined up in columns. This makes it easier to see which coefficients can be manipulated to create additive inverses. Once the equations are aligned, the next step is to choose a variable to eliminate. This choice often depends on the coefficients of the variables; it's usually easiest to eliminate the variable whose coefficients have a simple multiple relationship or are already close to being additive inverses.

Once a variable is chosen for elimination, the equations are multiplied by constants so that the coefficients of the chosen variable become additive inverses. This means that they have the same magnitude but opposite signs (e.g., 3 and -3). The multiplication must be applied to the entire equation, ensuring that the equality is maintained. This step requires careful attention to detail, as any error in multiplication will lead to an incorrect solution.

After the equations have been multiplied, they are added together. This is where the magic of the elimination method happens: the terms with the additive inverse coefficients cancel out, leaving an equation with only one variable. This resulting equation is typically a simple linear equation that can be solved using basic algebraic techniques. Once the value of one variable is found, it is substituted back into one of the original equations (or any equation in the process) to solve for the other variable. This process, known as back-substitution, completes the solution of the system.

It's important to note that the elimination method can also be used with more than two equations and variables. In such cases, the goal is to eliminate variables one at a time until the system is reduced to a set of equations that can be easily solved. This method is a fundamental tool in linear algebra and has wide applications in various fields, from engineering and physics to economics and computer science. A thorough understanding of the elimination method is crucial for anyone working with systems of linear equations.