Operations to Create Equivalent Systems

Oct 13, 2025 by thesweetdesigns.com 138 views

Which Operations Would Create an Equivalent System of Equations with Opposite Like Terms?

In algebra, a system of equations is a set of two or more equations that have the same variables. When we have a system of equations, our goal is to find the values of the variables that satisfy all the equations simultaneously. However, sometimes we may want to create a new system of equations that has the same solution as the original system, but with opposite like terms. In this article, we will explore the operations that can be performed on a system of equations to create an equivalent system with opposite like terms.

Understanding Like Terms

Before we dive into the operations, let's first understand what like terms are. Like terms are terms in an equation that have the same variable(s) raised to the same power. For example, in the equation 3x - 3y = 3, the terms 3x and -3y are like terms because they both have the variable x (or y) raised to the power of 1.

Operations to Create Equivalent Systems

Now that we understand like terms, let's explore the operations that can be performed on a system of equations to create an equivalent system with opposite like terms.

Multiplying Both Equations by a Constant

One way to create an equivalent system of equations with opposite like terms is to multiply both equations by a constant. This operation will multiply both like terms in each equation by the same constant, resulting in opposite like terms.

For example, consider the system of equations:

{ \begin{array}{l} 3x - 3y = 3 \\ 4x + 5y = 13 \end{array} \}

If we multiply both equations by 5, we get:

{ \begin{array}{l} 15x - 15y = 15 \\ 20x + 25y = 65 \end{array} \}

As we can see, the like terms in each equation have been multiplied by 5, resulting in opposite like terms.

Multiplying One Equation by a Constant and the Other Equation by a Different Constant

Another way to create an equivalent system of equations with opposite like terms is to multiply one equation by a constant and the other equation by a different constant. This operation will multiply the like terms in each equation by different constants, resulting in opposite like terms.

For example, consider the system of equations:

{ \begin{array}{l} 3x - 3y = 3 \\ 4x + 5y = 13 \end{array} \}

If we multiply the first equation by 5 and the second equation by 3, we get:

{ \begin{array}{l} 15x - 15y = 15 \\ 12x + 15y = 39 \end{array} \}

As we can see, the like terms in each equation have been multiplied by different constants, resulting in opposite like terms.

Adding or Subtracting the Equations

Another way to create an equivalent system of equations with opposite like terms is to add or subtract the equations. This operation will eliminate the like terms in the resulting equation, resulting in an equation with opposite like terms.

For example, consider the system of equations:

{ \begin{array}{l} 3x - 3y = 3 \\ 4x + 5y = 13 \end{array} \}

If we add the two equations, we get:

{ \begin{array}{l} 7x + 2y = 16 \end{array} \}

As we can see, the like terms in the original equations have been eliminated, resulting in an equation with opposite like terms.

Multiplying One Equation by a Negative Constant

Finally, another way to create an equivalent system of equations with opposite like terms is to multiply one equation by a negative constant. This operation will multiply the like terms in the equation by a negative constant, resulting in opposite like terms.

For example, consider the system of equations:

{ \begin{array}{l} 3x - 3y = 3 \\ 4x + 5y = 13 \end{array} \}

If we multiply the first equation by -1, we get:

{ \begin{array}{l} -3x + 3y = -3 \\ 4x + 5y = 13 \end{array} \}

As we can see, the like terms in the first equation have been multiplied by a negative constant, resulting in opposite like terms.

In conclusion, there are several operations that can be performed on a system of equations to create an equivalent system with opposite like terms. These operations include multiplying both equations by a constant, multiplying one equation by a constant and the other equation by a different constant, adding or subtracting the equations, and multiplying one equation by a negative constant. By understanding these operations, we can create new systems of equations that have the same solution as the original system, but with opposite like terms.

Problem 1

Consider the system of equations:

{ \begin{array}{l} 2x + 3y = 5 \\ 4x - 2y = 3 \end{array} \}

If we multiply both equations by 2, what is the resulting system of equations?

Solution

If we multiply both equations by 2, we get:

{ \begin{array}{l} 4x + 6y = 10 \\ 8x - 4y = 6 \end{array} \}

Problem 2

Consider the system of equations:

{ \begin{array}{l} x - 2y = 3 \\ 2x + 3y = 5 \end{array} \}

If we multiply the first equation by 2 and the second equation by 3, what is the resulting system of equations?

Solution

If we multiply the first equation by 2 and the second equation by 3, we get:

{ \begin{array}{l} 2x - 4y = 6 \\ 6x + 9y = 15 \end{array} \}

Problem 3

Consider the system of equations:

{ \begin{array}{l} x + 2y = 3 \\ 2x - 3y = 5 \end{array} \}

If we add the two equations, what is the resulting equation?

Solution

If we add the two equations, we get:

{ \begin{array}{l} 4x - y = 8 \end{array} \}

Problem 4

Consider the system of equations:

{ \begin{array}{l} 2x + 3y = 5 \\ 4x - 2y = 3 \end{array} \}

If we multiply the first equation by -1, what is the resulting system of equations?

Solution

If we multiply the first equation by -1, we get:

{ \begin{array}{l} -2x - 3y = -5 \\ 4x - 2y = 3 \end{array} \}

In conclusion, creating an equivalent system of equations with opposite like terms is a useful technique in algebra. By understanding the operations that can be performed on a system of equations, we can create new systems of equations that have the same solution as the original system, but with opposite like terms. This technique can be useful in a variety of applications, including solving systems of equations and creating new systems of equations with specific properties.
Q&A: Creating Equivalent Systems of Equations with Opposite Like Terms

In our previous article, we explored the operations that can be performed on a system of equations to create an equivalent system with opposite like terms. In this article, we will answer some frequently asked questions about creating equivalent systems of equations with opposite like terms.

Q: What is the purpose of creating an equivalent system of equations with opposite like terms?

A: The purpose of creating an equivalent system of equations with opposite like terms is to simplify the system of equations and make it easier to solve. By creating an equivalent system with opposite like terms, we can eliminate the like terms in the resulting equation, making it easier to solve for the variables.

Q: How do I know if two equations are equivalent?

A: Two equations are equivalent if they have the same solution. In other words, if the two equations have the same values for the variables, then they are equivalent.

Q: What is the difference between an equivalent system of equations and a system of equations with opposite like terms?

A: An equivalent system of equations is a system of equations that has the same solution as the original system. A system of equations with opposite like terms is a system of equations that has the same solution as the original system, but with opposite like terms.

Q: Can I create an equivalent system of equations with opposite like terms by adding or subtracting the equations?

A: Yes, you can create an equivalent system of equations with opposite like terms by adding or subtracting the equations. However, this method is not always the most efficient way to create an equivalent system with opposite like terms.

Q: Can I create an equivalent system of equations with opposite like terms by multiplying one equation by a negative constant?

A: Yes, you can create an equivalent system of equations with opposite like terms by multiplying one equation by a negative constant. This method is often the most efficient way to create an equivalent system with opposite like terms.

Q: How do I know if an equation has opposite like terms?

A: An equation has opposite like terms if the coefficients of the like terms are additive inverses. In other words, if the coefficients of the like terms are the same but have opposite signs, then the equation has opposite like terms.

Q: Can I create an equivalent system of equations with opposite like terms by multiplying both equations by a constant?

A: Yes, you can create an equivalent system of equations with opposite like terms by multiplying both equations by a constant. However, this method is not always the most efficient way to create an equivalent system with opposite like terms.

Q: Can I create an equivalent system of equations with opposite like terms by using a combination of the above methods?

A: Yes, you can create an equivalent system of equations with opposite like terms by using a combination of the above methods. For example, you can multiply one equation by a constant and the other equation by a different constant, or you can add or subtract the equations and then multiply one equation by a negative constant.

Q: How do I know if a system of equations has an equivalent system with opposite like terms?

A: A system of equations has an equivalent system with opposite like terms if the system can be transformed into an equivalent system with opposite like terms by using the above methods.

Q: Can I use a calculator to create an equivalent system of equations with opposite like terms?

A: Yes, you can use a calculator to create an equivalent system of equations with opposite like terms. However, it is often more efficient to use the above methods to create an equivalent system with opposite like terms.

Problem 1

Consider the system of equations:

{ \begin{array}{l} 2x + 3y = 5 \\ 4x - 2y = 3 \end{array} \}

If we multiply both equations by 2, what is the resulting system of equations?

Solution

If we multiply both equations by 2, we get:

{ \begin{array}{l} 4x + 6y = 10 \\ 8x - 4y = 6 \end{array} \}

Problem 2

Consider the system of equations:

{ \begin{array}{l} x - 2y = 3 \\ 2x + 3y = 5 \end{array} \}

If we multiply the first equation by 2 and the second equation by 3, what is the resulting system of equations?

Solution

If we multiply the first equation by 2 and the second equation by 3, we get:

{ \begin{array}{l} 2x - 4y = 6 \\ 6x + 9y = 15 \end{array} \}

Problem 3

Consider the system of equations:

{ \begin{array}{l} x + 2y = 3 \\ 2x - 3y = 5 \end{array} \}

If we add the two equations, what is the resulting equation?

Solution

If we add the two equations, we get:

{ \begin{array}{l} 4x - y = 8 \end{array} \}

Problem 4

Consider the system of equations:

{ \begin{array}{l} 2x + 3y = 5 \\ 4x - 2y = 3 \end{array} \}

If we multiply the first equation by -1, what is the resulting system of equations?

Solution

If we multiply the first equation by -1, we get:

{ \begin{array}{l} -2x - 3y = -5 \\ 4x - 2y = 3 \end{array} \}