Solve The Following System Of Equations Using The Elimination Method.$\[ 6x - 5y = -1 \\]$\[ -6x - 4y = 10 \\]A) \[$(5, -5)\$\]B) \[$(-2, 2)\$\]C) \[$(-1, -1)\$\]D) \[$(1, 1)\$\]

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Introduction

In algebra, a system of equations is a set of two or more equations that have the same variables. Solving a system of equations involves finding the values of the variables that satisfy all the equations in the system. There are several methods to solve a system of equations, including the substitution method, the elimination method, and the graphing method. In this article, we will focus on solving a system of equations using the elimination method.

What is the Elimination Method?

The elimination method is a technique used to solve a system of equations by adding or subtracting the equations to eliminate one of the variables. This method is useful when the coefficients of the variables in the equations are additive inverses. The elimination method involves the following steps:

  1. Multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same.
  2. Add or subtract the equations to eliminate the variable.
  3. Solve for the remaining variable.
  4. Substitute the value of the remaining variable into one of the original equations to solve for the other variable.

Step 1: Multiply the Equations by Necessary Multiples

To solve the given system of equations using the elimination method, we need to multiply the equations by necessary multiples such that the coefficients of the variable to be eliminated are the same.

The given system of equations is:

6x−5y=−1{ 6x - 5y = -1 }

−6x−4y=10{ -6x - 4y = 10 }

We can multiply the first equation by 1 and the second equation by 1.5 to make the coefficients of x the same.

6x−5y=−1{ 6x - 5y = -1 }

−9x−6y=15{ -9x - 6y = 15 }

Step 2: Add or Subtract the Equations to Eliminate the Variable

Now that the coefficients of x are the same, we can add the equations to eliminate the variable x.

(6x−5y)+(−9x−6y)=(−1)+15{ (6x - 5y) + (-9x - 6y) = (-1) + 15 }

−3x−11y=14{ -3x - 11y = 14 }

However, we want to eliminate x, so we need to multiply the first equation by 3 and the second equation by 2 to make the coefficients of x the same.

18x−15y=−3{ 18x - 15y = -3 }

−12x−8y=20{ -12x - 8y = 20 }

Now, we can add the equations to eliminate the variable x.

(18x−15y)+(−12x−8y)=(−3)+20{ (18x - 15y) + (-12x - 8y) = (-3) + 20 }

6x−23y=17{ 6x - 23y = 17 }

Step 3: Solve for the Remaining Variable

Now that we have eliminated the variable x, we can solve for the remaining variable y.

We have the equation:

6x−23y=17{ 6x - 23y = 17 }

However, we don't have the value of x. We need to go back to the original equations and eliminate x from one of them.

Let's go back to the original equations:

6x−5y=−1{ 6x - 5y = -1 }

−6x−4y=10{ -6x - 4y = 10 }

We can multiply the first equation by 1 and the second equation by 1 to make the coefficients of x the same.

6x−5y=−1{ 6x - 5y = -1 }

−6x−4y=10{ -6x - 4y = 10 }

Now, we can add the equations to eliminate the variable x.

(6x−5y)+(−6x−4y)=(−1)+10{ (6x - 5y) + (-6x - 4y) = (-1) + 10 }

−9y=9{ -9y = 9 }

Now, we can solve for y.

y=−1{ y = -1 }

Step 4: Substitute the Value of the Remaining Variable into One of the Original Equations to Solve for the Other Variable

Now that we have the value of y, we can substitute it into one of the original equations to solve for the other variable x.

Let's substitute y = -1 into the first original equation:

6x−5y=−1{ 6x - 5y = -1 }

6x−5(−1)=−1{ 6x - 5(-1) = -1 }

6x+5=−1{ 6x + 5 = -1 }

6x=−6{ 6x = -6 }

x=−1{ x = -1 }

Conclusion

In this article, we solved a system of equations using the elimination method. We multiplied the equations by necessary multiples to make the coefficients of the variable to be eliminated the same, added or subtracted the equations to eliminate the variable, solved for the remaining variable, and substituted the value of the remaining variable into one of the original equations to solve for the other variable.

The final answer is:

x=−1{ x = -1 }

y=−1{ y = -1 }

The correct answer is:

C) [$(-1, -1)$]

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Q: What is the elimination method?

A: The elimination method is a technique used to solve a system of equations by adding or subtracting the equations to eliminate one of the variables. This method is useful when the coefficients of the variables in the equations are additive inverses.

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 equations. If the coefficients of one variable are additive inverses, you can eliminate that variable first. If not, you need to multiply the equations by necessary multiples to make the coefficients of one variable the same.

Q: What if the coefficients of the variables are not additive inverses?

A: If the coefficients of the variables are not additive inverses, you need to multiply the equations by necessary multiples to make the coefficients of one variable the same. This will allow you to eliminate the variable by adding or subtracting the equations.

Q: Can I use the elimination method to solve a system of equations with three or more variables?

A: Yes, you can use the elimination method to solve a system of equations with three or more variables. However, you need to eliminate one variable at a time, and then use the resulting equation to eliminate another variable.

Q: How do I know if the system of equations has a unique solution, no solution, or infinitely many solutions?

A: To determine if the system of equations has a unique solution, no solution, or infinitely many solutions, you need to check the number of solutions to the system. If the system has a unique solution, it means that there is only one set of values that satisfies all the equations. If the system has no solution, it means that there is no set of values that satisfies all the equations. If the system has infinitely many solutions, it means that there are many sets of values that satisfy all the equations.

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

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

  • Not multiplying the equations by necessary multiples to make the coefficients of one variable the same.
  • Not adding or subtracting the equations correctly to eliminate the variable.
  • Not solving for the remaining variable correctly.
  • Not checking the number of solutions to the system.

Q: Can I use the elimination method to solve a system of equations with fractions or decimals?

A: Yes, you can use the elimination method to solve a system of equations with fractions or decimals. However, you need to follow the same steps as before, and make sure to multiply the equations by necessary multiples to eliminate the variable.

Q: How do I know if the elimination method is the best method to use to solve a system of equations?

A: To determine if the elimination method is the best method to use to solve a system of equations, you need to consider the following factors:

  • The complexity of the equations.
  • The number of variables in the system.
  • The coefficients of the variables in the equations.
  • The number of solutions to the system.

If the system has a simple structure and the coefficients of the variables are additive inverses, the elimination method may be the best method to use. However, if the system is complex or has many variables, another method such as the substitution method or the graphing method may be more suitable.

Conclusion

In this article, we answered some frequently asked questions about solving systems of equations using the elimination method. We discussed the steps involved in the elimination method, common mistakes to avoid, and how to determine if the elimination method is the best method to use to solve a system of equations.