Solve The System Using Addition. Use Pencil And Paper.Explain Why The Addition Method Is A Good Choice For Solving The System. If You Wanted To Solve For $x$ First, Is The Addition Method Still A Good Choice? Explain.$\[ \begin{array}{r} x

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Introduction

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In this article, we will explore the addition method for solving a system of linear equations. We will examine why the addition method is a good choice for solving the system and discuss the advantages of using this method. Additionally, we will consider the scenario where we want to solve for $x$ first and determine if the addition method is still a good choice.

The System of Linear Equations

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The system of linear equations we will be working with is:

$$\begin{array}{r} x + 2y = 6 \\ 3x - 2y = -3 \end{array}$$

The Addition Method

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The addition method involves adding the two equations together to eliminate one of the variables. In this case, we can add the two equations together to eliminate the variable $y$.

Step 1: Add the Two Equations Together

To add the two equations together, we need to add the corresponding terms. This gives us:

$$\begin{array}{r} x + 2y + 3x - 2y = 6 + (-3) \\ 4x = 3 \end{array}$$

Step 2: Solve for $x$

Now that we have eliminated the variable $y$, we can solve for $x$. To do this, we need to isolate $x$ on one side of the equation. We can do this by dividing both sides of the equation by 4:

$$\begin{array}{r} \frac{4x}{4} = \frac{3}{4} \\ x = \frac{3}{4} \end{array}$$

Step 3: Substitute $x$ into One of the Original Equations

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

$$\begin{array}{r} \frac{3}{4} + 2y = 6 \\ 2y = 6 - \frac{3}{4} \\ 2y = \frac{21}{4} \\ y = \frac{21}{8} \end{array}$$

Why the Addition Method is a Good Choice

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The addition method is a good choice for solving this system of linear equations because it allows us to eliminate one of the variables by adding the two equations together. This makes it easier to solve for the other variable. Additionally, the addition method is a simple and straightforward method that is easy to understand and apply.

Is the Addition Method Still a Good Choice if We Want to Solve for $x$ First?

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If we want to solve for $x$ first, the addition method is still a good choice. However, we need to be careful when adding the two equations together. If we add the two equations together without considering the signs of the coefficients, we may end up with an incorrect equation.

For example, if we add the two equations together without considering the signs of the coefficients, we get:

$$\begin{array}{r} x + 2y + 3x - 2y = 6 + (-3) \\ 4x = 3 \end{array}$$

However, if we consider the signs of the coefficients, we get:

$$\begin{array}{r} x + 2y + 3x - 2y = 6 + (-3) \\ (1 + 3)x + (2 - 2)y = 6 + (-3) \\ 4x = 3 \end{array}$$

As we can see, considering the signs of the coefficients is important when adding the two equations together. If we don't consider the signs of the coefficients, we may end up with an incorrect equation.

Conclusion

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In conclusion, the addition method is a good choice for solving this system of linear equations because it allows us to eliminate one of the variables by adding the two equations together. This makes it easier to solve for the other variable. Additionally, the addition method is a simple and straightforward method that is easy to understand and apply. If we want to solve for $x$ first, the addition method is still a good choice, but we need to be careful when adding the two equations together and consider the signs of the coefficients.

Example Problems

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Problem 1

Solve the system of linear equations using the addition method:

$$\begin{array}{r} 2x + 3y = 7 \\ x - 2y = -3 \end{array}$$

Solution

To solve this system of linear equations, we can add the two equations together to eliminate the variable $y$:

$$\begin{array}{r} 2x + 3y + x - 2y = 7 + (-3) \\ 3x + y = 4 \end{array}$$

Now that we have eliminated the variable $y$, we can solve for $x$. To do this, we need to isolate $x$ on one side of the equation. We can do this by subtracting $y$ from both sides of the equation:

$$\begin{array}{r} 3x = 4 - y \\ x = \frac{4 - y}{3} \end{array}$$

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

$$\begin{array}{r} 2\left(\frac{4 - y}{3}\right) + 3y = 7 \\ \frac{8 - 2y}{3} + 3y = 7 \\ 8 - 2y + 9y = 21 \\ 7y = 13 \\ y = \frac{13}{7} \end{array}$$

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

$$\begin{array}{r} 2x + 3\left(\frac{13}{7}\right) = 7 \\ 2x + \frac{39}{7} = 7 \\ 14x + 39 = 49 \\ 14x = 10 \\ x = \frac{5}{7} \end{array}$$

Problem 2

Solve the system of linear equations using the addition method:

$$\begin{array}{r} x + 2y = 6 \\ 2x + 3y = 11 \end{array}$$

Solution

To solve this system of linear equations, we can add the two equations together to eliminate the variable $y$:

$$\begin{array}{r} x + 2y + 2x + 3y = 6 + 11 \\ 3x + 5y = 17 \end{array}$$

Now that we have eliminated the variable $y$, we can solve for $x$. To do this, we need to isolate $x$ on one side of the equation. We can do this by subtracting $5y$ from both sides of the equation:

$$\begin{array}{r} 3x = 17 - 5y \\ x = \frac{17 - 5y}{3} \end{array}$$

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

$$\begin{array}{r} \left(\frac{17 - 5y}{3}\right) + 2y = 6 \\ 17 - 5y + 6y = 18 \\ y = \frac{1}{1} \\ y = 1 \end{array}$$

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

$$\begin{array}{r} x + 2(1) = 6 \\ x + 2 = 6 \\ x = 4 \end{array}$$

Final Answer

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The final answer is $\boxed{\left(\frac{3}{4}, \frac{21}{8}\right)}$.

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Introduction

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In our previous article, we explored the addition method for solving systems of linear equations. We discussed why the addition method is a good choice for solving these types of equations and provided examples of how to use this method. In this article, we will answer some frequently asked questions about solving systems of linear equations using addition.

Q&A

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

A: The addition method involves adding the two equations together to eliminate one of the variables. This makes it easier to solve for the other variable.

Q: Why is the addition method a good choice for solving systems of linear equations?

A: The addition method is a good choice for solving systems of linear equations because it allows us to eliminate one of the variables by adding the two equations together. This makes it easier to solve for the other variable.

Q: How do I know which variable to eliminate first?

A: To determine which variable to eliminate first, you need to look at the coefficients of the variables in the two equations. If the coefficients of one variable are the same, you can eliminate that variable first. If the coefficients of the variables are different, you can eliminate the variable with the smaller coefficient first.

Q: What if I want to solve for one variable first?

A: If you want to solve for one variable first, you can use the addition method to eliminate that variable. However, you need to be careful when adding the two equations together and consider the signs of the coefficients.

Q: Can I use the addition method to solve systems of linear equations with more than two variables?

A: Yes, you can use the addition method to solve systems of linear equations with more than two variables. However, you need to be careful when adding the equations together and consider the signs of the coefficients.

Q: What if I get a contradictory equation when using the addition method?

A: If you get a contradictory equation when using the addition method, it means that the system of linear equations has no solution. This can happen if the two equations are inconsistent.

Q: Can I use the addition method to solve systems of linear equations with fractions?

A: Yes, you can use the addition method to solve systems of linear equations with fractions. However, you need to be careful when adding the fractions together and consider the signs of the coefficients.

Example Problems

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Problem 1

Solve the system of linear equations using the addition method:

$$\begin{array}{r} 2x + 3y = 7 \\ x - 2y = -3 \end{array}$$

Solution

To solve this system of linear equations, we can add the two equations together to eliminate the variable $y$:

$$\begin{array}{r} 2x + 3y + x - 2y = 7 + (-3) \\ 3x + y = 4 \end{array}$$

Now that we have eliminated the variable $y$, we can solve for $x$. To do this, we need to isolate $x$ on one side of the equation. We can do this by subtracting $y$ from both sides of the equation:

$$\begin{array}{r} 3x = 4 - y \\ x = \frac{4 - y}{3} \end{array}$$

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

$$\begin{array}{r} 2\left(\frac{4 - y}{3}\right) + 3y = 7 \\ \frac{8 - 2y}{3} + 3y = 7 \\ 8 - 2y + 9y = 21 \\ 7y = 13 \\ y = \frac{13}{7} \end{array}$$

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

$$\begin{array}{r} 2x + 3\left(\frac{13}{7}\right) = 7 \\ 2x + \frac{39}{7} = 7 \\ 14x + 39 = 49 \\ 14x = 10 \\ x = \frac{5}{7} \end{array}$$

Problem 2

Solve the system of linear equations using the addition method:

$$\begin{array}{r} x + 2y = 6 \\ 2x + 3y = 11 \end{array}$$

Solution

To solve this system of linear equations, we can add the two equations together to eliminate the variable $y$:

$$\begin{array}{r} x + 2y + 2x + 3y = 6 + 11 \\ 3x + 5y = 17 \end{array}$$

Now that we have eliminated the variable $y$, we can solve for $x$. To do this, we need to isolate $x$ on one side of the equation. We can do this by subtracting $5y$ from both sides of the equation:

$$\begin{array}{r} 3x = 17 - 5y \\ x = \frac{17 - 5y}{3} \end{array}$$

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

$$\begin{array}{r} \left(\frac{17 - 5y}{3}\right) + 2y = 6 \\ 17 - 5y + 6y = 18 \\ y = \frac{1}{1} \\ y = 1 \end{array}$$

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

$$\begin{array}{r} x + 2(1) = 6 \\ x + 2 = 6 \\ x = 4 \end{array}$$

Final Answer

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The final answer is $\boxed{\left(\frac{3}{4}, \frac{21}{8}\right)}$.