Solve The System Of Equations Below Using Elimination By Addition.$\[ \begin{aligned} 5m - N &= 25 \\ m - 5n &= -19 \end{aligned} \\]Select The Correct Choice Below And, If Necessary, Fill In The Answer Box To Complete Your Choice.A. The

Introduction

In mathematics, solving systems of equations is a fundamental concept that involves finding the values of variables that satisfy multiple equations simultaneously. One of the methods used to solve systems of equations is the elimination by addition method. This method involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the other variable. In this article, we will use the elimination by addition method to solve a system of two linear equations.

The System of Equations

The system of equations we will be solving is given as:

{ \begin{aligned} 5m - n &= 25 \\ m - 5n &= -19 \end{aligned} \}

Our goal is to find the values of $m$ and $n$ that satisfy both equations.

Step 1: Multiply the Equations by Necessary Multiples

To eliminate one of the variables, we need to make the coefficients of either $m$ or $n$ the same in both equations. We can do this by multiplying the equations by necessary multiples.

Let's multiply the first equation by 1 and the second equation by 5. This will give us:

{ \begin{aligned} 5m - n &= 25 \\ 5(m - 5n) &= 5(-19) \end{aligned} \}

Simplifying the second equation, we get:

{ \begin{aligned} 5m - n &= 25 \\ 5m - 25n &= -95 \end{aligned} \}

Step 2: Add the Equations

Now that we have the equations in a form where the coefficients of $m$ are the same, we can add the equations to eliminate $m$. Adding the equations, we get:

{ \begin{aligned} (5m - n) + (5m - 25n) &= 25 + (-95) \\ 10m - 26n &= -70 \end{aligned} \}

Step 3: Solve for $n$

Now that we have eliminated $m$, we can solve for $n$. We can do this by isolating $n$ on one side of the equation. Adding $26n$ to both sides of the equation, we get:

{ \begin{aligned} 10m - 26n + 26n &= -70 + 26n \\ 10m &= 26n - 70 \end{aligned} \}

Dividing both sides of the equation by 10, we get:

{ \begin{aligned} \frac{10m}{10} &= \frac{26n - 70}{10} \\ m &= \frac{26n}{10} - \frac{70}{10} \\ m &= \frac{13n}{5} - 7 \end{aligned} \}

Step 4: Substitute $m$ into One of the Original Equations

Now that we have expressed $m$ in terms of $n$, we can substitute this expression into one of the original equations to solve for $n$. Let's substitute this expression into the first original equation:

{ \begin{aligned} 5m - n &= 25 \\ 5(\frac{13n}{5} - 7) - n &= 25 \end{aligned} \}

Simplifying the equation, we get:

{ \begin{aligned} \frac{65n}{5} - 35 - n &= 25 \\ \frac{65n}{5} - \frac{5n}{5} &= 25 + 35 \\ \frac{60n}{5} &= 60 \\ 60n &= 300 \end{aligned} \}

Dividing both sides of the equation by 60, we get:

{ \begin{aligned} \frac{60n}{60} &= \frac{300}{60} \\ n &= 5 \end{aligned} \}

Step 5: Substitute $n$ into the Expression for $m$

Now that we have found the value of $n$, we can substitute this value into the expression for $m$ to find the value of $m$:

{ \begin{aligned} m &= \frac{13n}{5} - 7 \\ m &= \frac{13(5)}{5} - 7 \\ m &= 13 - 7 \\ m &= 6 \end{aligned} \}

Conclusion

In this article, we used the elimination by addition method to solve a system of two linear equations. We multiplied the equations by necessary multiples, added the equations to eliminate one of the variables, solved for the other variable, and substituted the value of the other variable into the expression for the first variable to find the values of both variables. The values of $m$ and $n$ that satisfy both equations are $m = 6$ and $n = 5$.

Discussion

The elimination by addition method is a powerful tool for solving systems of equations. By adding or subtracting equations, we can eliminate one of the variables, making it easier to solve for the other variable. This method can be used to solve systems of linear equations with two or more variables. However, it is essential to choose the correct equations to add or subtract to eliminate the correct variable.

Example Problems

1. Solve the system of equations using elimination by addition:

{ \begin{aligned} 2x + 3y &= 7 \\ x - 2y &= -3 \end{aligned} \}

2. Solve the system of equations using elimination by addition:

{ \begin{aligned} x + 2y &= 4 \\ 3x - 2y &= 2 \end{aligned} \}

Solutions

1. The values of $x$ and $y$ that satisfy both equations are $x = 1$ and $y = 2$.

2. The values of $x$ and $y$ that satisfy both equations are $x = 2$ and $y = 1$.

References

1. "Algebra and Trigonometry" by Michael Sullivan
2. "College Algebra" by James Stewart
3. "Linear Algebra and Its Applications" by Gilbert Strang

Glossary

• Elimination by addition: A method used to solve systems of equations by adding or subtracting equations to eliminate one of the variables.
• System of equations: A set of two or more equations that involve two or more variables.
• Linear equation: An equation in which the highest power of the variable is 1.
• Variable: A symbol or expression that represents a value that can change.
  Solving Systems of Equations using Elimination by Addition: Q&A ================================================================

Introduction

In our previous article, we used the elimination by addition method to solve a system of two linear equations. In this article, we will answer some frequently asked questions about solving systems of equations using elimination by addition.

Q: What is the elimination by addition method?

A: The elimination by addition method is a technique used to solve systems of equations by adding or subtracting equations to eliminate one of the variables. This method involves making the coefficients of either the variable to be eliminated or the variable to be solved the same in both equations.

Q: How do I choose which equations to add or subtract?

A: To choose which equations to add or subtract, you need to make the coefficients of either the variable to be eliminated or the variable to be solved the same in both equations. You can do this by multiplying the equations by necessary multiples.

Q: What if the coefficients of the variables are not the same in both equations?

A: If the coefficients of the variables are not the same in both equations, you can multiply the equations by necessary multiples to make the coefficients the same. For example, if the coefficients of $x$ are 2 and 3 in the two equations, you can multiply the first equation by 3 and the second equation by 2 to make the coefficients of $x$ the same.

Q: How do I eliminate one of the variables?

A: To eliminate one of the variables, you need to add or subtract the equations in such a way that the variable to be eliminated is eliminated. For example, if you want to eliminate the variable $x$, you can add the two equations together to eliminate $x$.

Q: What if I have a system of three or more equations?

A: If you have a system of three or more equations, you can use the elimination by addition method to solve the system by adding or subtracting the equations in pairs to eliminate one of the variables at a time.

Q: Can I use the elimination by addition method to solve non-linear equations?

A: No, the elimination by addition method is only used to solve linear equations. If you have a system of non-linear equations, you will need to use a different method, such as substitution or graphing.

Q: What are some common mistakes to avoid when using the elimination by addition method?

A: Some common mistakes to avoid when using the elimination by addition method include:

• Not making the coefficients of the variables the same in both equations
• Not adding or subtracting the equations correctly to eliminate the variable
• Not checking the solution to make sure it satisfies both equations

Q: How do I check the solution to make sure it satisfies both equations?

A: To check the solution, you need to substitute the values of the variables into both equations and make sure they are true. If the solution satisfies both equations, then it is the correct solution.

Conclusion

In this article, we have answered some frequently asked questions about solving systems of equations using elimination by addition. We have discussed how to choose which equations to add or subtract, how to eliminate one of the variables, and how to check the solution to make sure it satisfies both equations. By following these steps, you can use the elimination by addition method to solve systems of linear equations.

Example Problems

1. Solve the system of equations using elimination by addition:

{ \begin{aligned} 2x + 3y &= 7 \\ x - 2y &= -3 \end{aligned} \}

2. Solve the system of equations using elimination by addition:

{ \begin{aligned} x + 2y &= 4 \\ 3x - 2y &= 2 \end{aligned} \}

Solutions

1. The values of $x$ and $y$ that satisfy both equations are $x = 1$ and $y = 2$.

2. The values of $x$ and $y$ that satisfy both equations are $x = 2$ and $y = 1$.

References

1. "Algebra and Trigonometry" by Michael Sullivan
2. "College Algebra" by James Stewart
3. "Linear Algebra and Its Applications" by Gilbert Strang

Glossary

• Elimination by addition: A method used to solve systems of equations by adding or subtracting equations to eliminate one of the variables.
• System of equations: A set of two or more equations that involve two or more variables.
• Linear equation: An equation in which the highest power of the variable is 1.
• Variable: A symbol or expression that represents a value that can change.