Eliminating X Variable Constants In System Of Equations 4x + 5y = 62 And 5x - 3y = 22

Jul 13, 2025 by ADMIN 86 views

Eliminating the x Variable in a System of Equations

Introduction

In the realm of mathematics, particularly in the study of linear algebra, systems of equations play a pivotal role. These systems, composed of two or more equations with shared variables, often arise in various real-world applications, ranging from engineering and physics to economics and computer science. A fundamental task in dealing with systems of equations is finding the values of the variables that satisfy all equations simultaneously. This process, known as solving the system, can be achieved through a variety of techniques, one of which involves eliminating variables.

Variable elimination is a powerful method that simplifies the system by strategically manipulating the equations to remove one or more variables. This reduction in the number of variables makes the system easier to solve, often leading to a straightforward solution for the remaining variables. One common approach to variable elimination is through multiplication by a constant. By multiplying one or both equations by carefully chosen constants, we can create coefficients for a specific variable that are either equal or additive inverses. This allows us to eliminate the variable by either subtracting or adding the equations, resulting in a new equation with fewer variables.

In this article, we will delve into the specific problem of eliminating the 'x' variable from a given system of equations. We will explore the underlying principles behind this technique, the steps involved in determining the appropriate constants, and the application of these constants to achieve the desired elimination. By understanding these concepts, you will gain a valuable tool for solving systems of equations and tackling a wide range of mathematical problems.

The System of Equations

Let us consider the following system of equations:

Equation 1: $4x + 5y = 62$

Equation 2: $5x - 3y = 22$

Our goal is to determine the constants by which each equation can be multiplied to eliminate the 'x' variable. This means we want to manipulate the equations so that the coefficients of 'x' become either equal or additive inverses. By doing so, we can then subtract or add the equations to eliminate 'x', leaving us with a single equation in terms of 'y'. This simplified equation can then be easily solved for 'y', and the value of 'y' can be substituted back into either of the original equations to find the value of 'x'.

Finding the Constants

To eliminate the 'x' variable, we need to find constants that, when multiplied by the respective equations, will result in either equal or additive inverse coefficients for 'x'. In this case, the coefficients of 'x' are 4 and 5. A common approach is to find the least common multiple (LCM) of these coefficients. The LCM of 4 and 5 is 20. Therefore, we aim to make the coefficients of 'x' in both equations equal to 20 or -20.

To achieve this, we can multiply Equation 1 by 5 and Equation 2 by -4. This will result in the following modifications:

Equation 1 multiplied by 5: $5(4x + 5y) = 5(62)$ which simplifies to $20x + 25y = 310$

Equation 2 multiplied by -4: $-4(5x - 3y) = -4(22)$ which simplifies to $-20x + 12y = -88$

Notice that the coefficients of 'x' are now 20 and -20, which are additive inverses. This sets the stage for eliminating 'x' by adding the two equations together.

Step-by-Step Guide to Finding Constants

Identify the coefficients: Begin by identifying the coefficients of the variable you want to eliminate (in this case, 'x') in both equations.
Find the Least Common Multiple (LCM): Determine the LCM of these coefficients. The LCM is the smallest positive integer that is divisible by both coefficients.
Calculate the multipliers: For each equation, divide the LCM by the coefficient of the variable you want to eliminate. This will give you the constant by which you need to multiply the equation.
Choose the sign: Decide whether you want the coefficients of the variable to be equal or additive inverses. If you want them to be equal, simply use the constants you calculated in the previous step. If you want them to be additive inverses, multiply one of the constants by -1.

By following these steps, you can systematically determine the constants needed to eliminate a variable in a system of equations.

Applying the Constants

As we determined in the previous section, we need to multiply Equation 1 by 5 and Equation 2 by -4 to eliminate the 'x' variable. Let's apply these constants and see the result.

Multiplying Equation 1 by 5:

$5(4x + 5y) = 5(62)$

$20x + 25y = 310$

Multiplying Equation 2 by -4:

$-4(5x - 3y) = -4(22)$

$-20x + 12y = -88$

Now, we have two new equations:

Equation 3: $20x + 25y = 310$

Equation 4: $-20x + 12y = -88$

Notice that the coefficients of 'x' in Equation 3 and Equation 4 are 20 and -20, respectively. These are additive inverses, which means that when we add the two equations together, the 'x' terms will cancel out.

Adding Equation 3 and Equation 4:

$(20x + 25y) + (-20x + 12y) = 310 + (-88)$

$20x - 20x + 25y + 12y = 310 - 88$

$37y = 222$

Now we have a single equation with only one variable, 'y'. This equation can be easily solved for 'y'.

Solving for y

To solve for 'y' in the equation $37y = 222$ , we simply divide both sides of the equation by 37:

$y = 222 / 37$

$y = 6$

Therefore, the value of 'y' that satisfies the system of equations is 6. Now that we have found the value of 'y', we can substitute it back into either of the original equations to find the value of 'x'.

Solving for x

Let's substitute the value of 'y' (which is 6) into Equation 1:

$4x + 5y = 62$

$4x + 5(6) = 62$

$4x + 30 = 62$

Now, we need to isolate 'x'. First, subtract 30 from both sides of the equation:

$4x = 62 - 30$

$4x = 32$

Next, divide both sides of the equation by 4:

$x = 32 / 4$

$x = 8$

Therefore, the value of 'x' that satisfies the system of equations is 8.

Solution

We have successfully eliminated the 'x' variable from the system of equations by multiplying Equation 1 by 5 and Equation 2 by -4. This led us to a new equation with only 'y', which we solved to find $y = 6$ . Substituting this value back into one of the original equations allowed us to solve for 'x', resulting in $x = 8$ .

Therefore, the solution to the system of equations is $x = 8$ and $y = 6$ . This means that the point (8, 6) is the intersection of the two lines represented by the equations $4x + 5y = 62$ and $5x - 3y = 22$ .

Conclusion

In this article, we explored the technique of eliminating a variable in a system of equations by multiplying the equations by constants. We focused on eliminating the 'x' variable from the system:

$4x + 5y = 62$

$5x - 3y = 22$

We determined that multiplying Equation 1 by 5 and Equation 2 by -4 would eliminate the 'x' variable. This process led us to the solution $x = 8$ and $y = 6$ .

This method of variable elimination is a powerful tool in solving systems of equations. It allows us to simplify the system by reducing the number of variables, making it easier to find the solution. By understanding the principles behind this technique and the steps involved in finding the appropriate constants, you can confidently tackle a wide range of mathematical problems involving systems of equations.

Remember, the key to successful variable elimination lies in finding the LCM of the coefficients of the variable you want to eliminate and then choosing the constants that will make the coefficients either equal or additive inverses. With practice, this technique will become a valuable asset in your mathematical toolkit.

This article has provided a comprehensive guide to eliminating the 'x' variable in a specific system of equations. However, the principles and techniques discussed can be applied to eliminate any variable in any system of equations. By mastering this skill, you will enhance your problem-solving abilities and gain a deeper understanding of linear algebra and its applications.