Solving Systems of Equations using Elimination

Oct 13, 2025 by vinooalluri.com 84 views

Introduction

Mathematics is a subject that requires practice and problem-solving skills to master. One of the essential topics in mathematics is solving systems of equations. In this article, we will focus on solving systems of equations using the elimination method. We will provide step-by-step solutions to three different systems of equations, and discuss the importance of practice and problem-solving skills in mathematics.

What is a System of Equations?

A system of equations is a set of two or more equations that contain the same variables. In other words, a system of equations is a collection of equations that are related to each other through the variables. For example, consider the following system of equations:

{ \begin{align*} x + y &= 4 \\ 2x - 3y &= -1 \end{align*} \}

This is a system of two equations with two variables, x and y. To solve this system, we need to find the values of x and y that satisfy both equations.

Solving Systems of Equations using Elimination

The elimination method is a popular method for solving systems of equations. This method involves adding or subtracting equations to eliminate one of the variables. To use the elimination method, we need to follow these steps:

Write down the system of equations: Write down the system of equations that we want to solve.
Identify the variables: Identify the variables in the system of equations.
Choose a method: Choose a method to eliminate one of the variables. We can use either the addition or subtraction method.
Eliminate the variable: Eliminate one of the variables by adding or subtracting equations.
Solve for the remaining variable: Solve for the remaining variable.
Find the value of the eliminated variable: Find the value of the eliminated variable.

Solving System 7

Let's solve the following system of equations using the elimination method:

{ \begin{align*} 3x + 3y &= 27 \\ x - 3y &= -11 \end{align*} \}

To solve this system, we can use the elimination method. First, we need to identify the variables and choose a method to eliminate one of the variables. In this case, we can eliminate the variable y by adding the two equations.

Step 1: Add the two equations

{ \begin{align*} (3x + 3y) + (x - 3y) &= 27 + (-11) \\ 4x &= 16 \end{align*} \}

Step 2: Solve for x

{ \begin{align*} x &= \frac{16}{4} \\ x &= 4 \end{align*} \}

Step 3: Find the value of y

Now that we have the value of x, we can substitute it into one of the original equations to find the value of y. Let's substitute x = 4 into the first equation:

{ \begin{align*} 3(4) + 3y &= 27 \\ 12 + 3y &= 27 \\ 3y &= 15 \\ y &= 5 \end{align*} \}

Therefore, the solution to the system is x = 4 and y = 5.

Solving System 8

Let's solve the following system of equations using the elimination method:

{ -x + 5y = 13 \}

Step 1: Multiply the equation by -1

{ \begin{align*} -x + 5y &= 13 \\ (-1)(-x + 5y) &= (-1)(13) \\ x - 5y &= -13 \end{align*} \}

Step 2: Solve for y

Now that we have the equation x - 5y = -13, we can solve for y. However, we need another equation to solve for y. Let's assume that the other equation is x + 5y = 23.

Step 3: Add the two equations

{ \begin{align*} (x - 5y) + (x + 5y) &= -13 + 23 \\ 2x &= 10 \\ x &= 5 \end{align*} \}

Step 4: Find the value of y

Now that we have the value of x, we can substitute it into one of the original equations to find the value of y. Let's substitute x = 5 into the equation x + 5y = 23:

{ \begin{align*} 5 + 5y &= 23 \\ 5y &= 18 \\ y &= \frac{18}{5} \\ y &= 3.6 \end{align*} \}

Therefore, the solution to the system is x = 5 and y = 3.6.

Solving System 9

Let's solve the following system of equations using the elimination method:

{ \begin{align*} 2x + 4y &= 22 \\ x - y &= 3 \end{align*} \}

Step 1: Multiply the second equation by 2

{ \begin{align*} 2x - 2y &= 6 \end{align*} \}

Step 2: Add the two equations

{ \begin{align*} (2x + 4y) + (2x - 2y) &= 22 + 6 \\ 4x + 2y &= 28 \end{align*} \}

Step 3: Solve for x

Now that we have the equation 4x + 2y = 28, we can solve for x. However, we need another equation to solve for x. Let's assume that the other equation is x - y = 3.

Step 4: Add the two equations

{ \begin{align*} (4x + 2y) + (x - y) &= 28 + 3 \\ 5x + y &= 31 \end{align*} \}

Step 5: Solve for x

Now that we have the equation 5x + y = 31, we can solve for x. However, we need another equation to solve for x. Let's assume that the other equation is x - y = 3.

Step 6: Add the two equations

{ \begin{align*} (5x + y) + (x - y) &= 31 + 3 \\ 6x &= 34 \\ x &= \frac{34}{6} \\ x &= 5.67 \end{align*} \}

Step 7: Find the value of y

Now that we have the value of x, we can substitute it into one of the original equations to find the value of y. Let's substitute x = 5.67 into the equation x - y = 3:

{ \begin{align*} 5.67 - y &= 3 \\ -y &= -2.67 \\ y &= 2.67 \end{align*} \}

Therefore, the solution to the system is x = 5.67 and y = 2.67.

Conclusion

Solving systems of equations using the elimination method requires practice and problem-solving skills. In this article, we have provided step-by-step solutions to three different systems of equations. We have also discussed the importance of practice and problem-solving skills in mathematics. By practicing and solving systems of equations, students can develop their problem-solving skills and become proficient in mathematics.

Practice Problems

Here are some practice problems to help you develop your problem-solving skills:

Solve the following system of equations using the elimination method:

{ \begin{align*} 2x + 3y &= 12 \\ x - 2y &= -3 \end{align*} \}

Solve the following system of equations using the elimination method:

{ \begin{align*} x + 2y &= 7 \\ 3x - 2y &= 13 \end{align*} \}

Solve the following system of equations using the elimination method:

{ \begin{align*} 4x + 3y &= 25 \\ 2x - 3y &= -11 \end{align*} \}

Discussion

Solving systems of equations using the elimination method is an essential skill in mathematics. By practicing and solving systems of equations, students can develop their problem-solving skills and become proficient in mathematics. In this article, we have provided step-by-step solutions to three different systems of equations and discussed the importance of practice and problem-solving skills in mathematics.

References

"Algebra and Trigonometry" by Michael Sullivan
"College Algebra" by James Stewart
"Mathematics for the Nonmathematician"
Mathematical Practice and Problem-Solving Exercises: Solving Systems of Equations ===========================================================

Q&A: Solving Systems of Equations using Elimination

Q: What is the elimination method for solving systems of equations?

A: The elimination method is a popular method for solving systems of equations. This method involves adding or subtracting equations to eliminate one of the variables. To use the elimination method, we need to follow these steps:

Write down the system of equations: Write down the system of equations that we want to solve.
Identify the variables: Identify the variables in the system of equations.
Choose a method: Choose a method to eliminate one of the variables. We can use either the addition or subtraction method.
Eliminate the variable: Eliminate one of the variables by adding or subtracting equations.
Solve for the remaining variable: Solve for the remaining variable.
Find the value of the eliminated variable: Find the value of the eliminated variable.

Q: How do I choose a method to eliminate one of the variables?

A: To choose a method to eliminate one of the variables, we need to look at the coefficients of the variables in the system of equations. We can use either the addition or subtraction method to eliminate one of the variables. For example, if we have the following system of equations:

{ \begin{align*} 2x + 3y &= 12 \\ x - 2y &= -3 \end{align*} \}

We can eliminate the variable x by multiplying the second equation by 2 and then adding the two equations.

Q: What is the difference between the addition and subtraction methods?

A: The addition and subtraction methods are two different methods for eliminating one of the variables in a system of equations. The addition method involves adding two equations to eliminate one of the variables, while the subtraction method involves subtracting two equations to eliminate one of the variables.

For example, if we have the following system of equations:

{ \begin{align*} 2x + 3y &= 12 \\ x - 2y &= -3 \end{align*} \}

We can use the addition method to eliminate the variable x by multiplying the second equation by 2 and then adding the two equations:

{ \begin{align*} (2x + 3y) + (2x - 4y) &= 12 + (-6) \\ 4x - y &= 6 \end{align*} \}

Alternatively, we can use the subtraction method to eliminate the variable x by subtracting the second equation from the first equation:

{ \begin{align*} (2x + 3y) - (x - 2y) &= 12 - (-3) \\ x + 5y &= 15 \end{align*} \}

Q: How do I solve for the remaining variable?

A: To solve for the remaining variable, we need to isolate the variable in one of the equations. We can do this by adding or subtracting the same value to both sides of the equation.

For example, if we have the following equation:

{ \begin{align*} 4x - y &= 6 \end{align*} \}

We can solve for x by adding y to both sides of the equation:

{ \begin{align*} 4x - y + y &= 6 + y \\ 4x &= 6 + y \end{align*} \}

Alternatively, we can solve for x by subtracting y from both sides of the equation:

{ \begin{align*} 4x - y - y &= 6 - y \\ 4x &= 6 - 2y \end{align*} \}

Q: How do I find the value of the eliminated variable?

A: To find the value of the eliminated variable, we need to substitute the value of the remaining variable into one of the original equations.

For example, if we have the following system of equations:

{ \begin{align*} 2x + 3y &= 12 \\ x - 2y &= -3 \end{align*} \}

We can solve for x by using the elimination method:

{ \begin{align*} (2x + 3y) + (2x - 4y) &= 12 + (-6) \\ 4x - y &= 6 \end{align*} \}

We can then solve for x by adding y to both sides of the equation:

{ \begin{align*} 4x - y + y &= 6 + y \\ 4x &= 6 + y \end{align*} \}

We can then substitute the value of x into one of the original equations to find the value of y:

{ \begin{align*} 2x + 3y &= 12 \\ 2(4) + 3y &= 12 \\ 8 + 3y &= 12 \\ 3y &= 4 \\ y &= \frac{4}{3} \end{align*} \}

Therefore, the value of the eliminated variable y is 4/3.

Conclusion

Solving systems of equations using the elimination method requires practice and problem-solving skills. By following the steps outlined in this article, you can develop your problem-solving skills and become proficient in mathematics. Remember to choose a method to eliminate one of the variables, solve for the remaining variable, and find the value of the eliminated variable.

Practice Problems

Here are some practice problems to help you develop your problem-solving skills:

Solve the following system of equations using the elimination method:

{ \begin{align*} 2x + 3y &= 12 \\ x - 2y &= -3 \end{align*} \}

Solve the following system of equations using the elimination method:

{ \begin{align*} x + 2y &= 7 \\ 3x - 2y &= 13 \end{align*} \}

Solve the following system of equations using the elimination method:

{ \begin{align*} 4x + 3y &= 25 \\ 2x - 3y &= -11 \end{align*} \}

Discussion

Solving systems of equations using the elimination method is an essential skill in mathematics. By practicing and solving systems of equations, you can develop your problem-solving skills and become proficient in mathematics. Remember to choose a method to eliminate one of the variables, solve for the remaining variable, and find the value of the eliminated variable.

References

"Algebra and Trigonometry" by Michael Sullivan
"College Algebra" by James Stewart
"Mathematics for the Nonmathematician"