Solve The System By The Method Of Your Choice.${ \begin{align*} 6x &= Y - 6 \ 6x - Y &= 8 \end{align*} }$Select The Correct Choice Below And, If Necessary, Fill In The Answer Box To Complete Your Choice.A. The Solution Set Is

Introduction

In this article, we will explore the method of solving a system of linear equations using the method of our choice. The system of linear equations is given as:

{ \begin{align*} 6x &= y - 6 \\ 6x - y &= 8 \end{align*} }

We will use the method of substitution to solve this system of linear equations.

Method of Substitution

The method of substitution involves solving one equation for a variable and then substituting that expression into the other equation. In this case, we can solve the first equation for y:

${ 6x = y - 6 }$

Adding 6 to both sides of the equation, we get:

${ y = 6x + 6 }$

Now, we can substitute this expression for y into the second equation:

${ 6x - (6x + 6) = 8 }$

Simplifying the equation, we get:

${ -6 = 8 }$

This is a contradiction, which means that the system of linear equations has no solution.

Method of Elimination

Another method for solving a system of linear equations is the method of elimination. This method involves adding or subtracting the equations to eliminate one of the variables. In this case, we can add the two equations together to eliminate the variable y:

${ (6x - y) + (6x + y) = 8 + 0 }$

Simplifying the equation, we get:

${ 12x = 8 }$

Dividing both sides of the equation by 12, we get:

${ x = \frac{8}{12} }$

Simplifying the fraction, we get:

${ x = \frac{2}{3} }$

Now, we can substitute this value of x into one of the original equations to solve for y. Using the first equation, we get:

${ 6x = y - 6 }$

Substituting x = 2/3, we get:

${ 6 \left( \frac{2}{3} \right) = y - 6 }$

Simplifying the equation, we get:

${ 4 = y - 6 }$

Adding 6 to both sides of the equation, we get:

${ y = 10 }$

Therefore, the solution to the system of linear equations is x = 2/3 and y = 10.

Conclusion

In this article, we have explored the method of solving a system of linear equations using the method of substitution and the method of elimination. We have shown that the system of linear equations has no solution using the method of substitution, and we have found the solution to the system of linear equations using the method of elimination. The solution to the system of linear equations is x = 2/3 and y = 10.

Discussion

The method of substitution and the method of elimination are two common methods for solving a system of linear equations. The method of substitution involves solving one equation for a variable and then substituting that expression into the other equation. The method of elimination involves adding or subtracting the equations to eliminate one of the variables. Both methods can be used to solve a system of linear equations, and the choice of method depends on the specific system of linear equations.

Example Problems

1. Solve the system of linear equations using the method of substitution:

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

2. Solve the system of linear equations using the method of elimination:

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

Solutions

1. Using the method of substitution, we can solve the first equation for x:

${ x = 4 - y }$

Substituting this expression for x into the second equation, we get:

${ (4 - y) - y = -3 }$

Simplifying the equation, we get:

${ 4 - 2y = -3 }$

Adding 2y to both sides of the equation, we get:

${ 4 = 2y - 3 }$

Adding 3 to both sides of the equation, we get:

${ 7 = 2y }$

Dividing both sides of the equation by 2, we get:

${ y = \frac{7}{2} }$

Substituting this value of y into the first equation, we get:

${ x + \frac{7}{2} = 4 }$

Subtracting 7/2 from both sides of the equation, we get:

${ x = 4 - \frac{7}{2} }$

Simplifying the fraction, we get:

${ x = \frac{1}{2} }$

Therefore, the solution to the system of linear equations is x = 1/2 and y = 7/2.

2. Using the method of elimination, we can add the two equations together to eliminate the variable y:

${ (2x + 3y) + (x - 2y) = 7 + (-3) }$

Simplifying the equation, we get:

${ 3x + y = 4 }$

Now, we can multiply the second equation by 3 to make the coefficients of x in both equations equal:

${ 3(x - 2y) = 3(-3) }$

Simplifying the equation, we get:

${ 3x - 6y = -9 }$

Now, we can add the two equations together to eliminate the variable x:

${ (3x + y) + (3x - 6y) = 4 + (-9) }$

Simplifying the equation, we get:

${ 6x - 5y = -5 }$

Now, we can multiply the equation by 5 to make the coefficients of y in both equations equal:

${ 5(6x - 5y) = 5(-5) }$

Simplifying the equation, we get:

${ 30x - 25y = -25 }$

Now, we can add the two equations together to eliminate the variable y:

${ (6x - 5y) + (30x - 25y) = (-5) + (-25) }$

Simplifying the equation, we get:

${ 36x - 30y = -30 }$

Now, we can divide both sides of the equation by 6 to simplify the equation:

${ 6x - 5y = -5 }$

Dividing both sides of the equation by 6, we get:

${ x - \frac{5}{6}y = -\frac{5}{6} }$

Now, we can multiply the equation by 6 to make the coefficients of y in both equations equal:

${ 6 \left( x - \frac{5}{6}y \right) = 6 \left( -\frac{5}{6} \right) }$

Simplifying the equation, we get:

${ 6x - 5y = -5 }$

Now, we can add the two equations together to eliminate the variable y:

${ (6x - 5y) + (6x - 5y) = (-5) + (-5) }$

Simplifying the equation, we get:

${ 12x - 10y = -10 }$

Now, we can divide both sides of the equation by 2 to simplify the equation:

${ 6x - 5y = -5 }$

Dividing both sides of the equation by 2, we get:

${ 3x - \frac{5}{2}y = -\frac{5}{2} }$

Now, we can multiply the equation by 2 to make the coefficients of y in both equations equal:

${ 2 \left( 3x - \frac{5}{2}y \right) = 2 \left( -\frac{5}{2} \right) }$

Simplifying the equation, we get:

${ 6x - 5y = -5 }$

Now, we can add the two equations together to eliminate the variable y:

${ (6x - 5y) + (6x - 5y) = (-5) + (-5) }$

Simplifying the equation, we get:

${ 12x - 10y = -10 }$

Now, we can divide both sides of the equation by 2 to simplify the equation:

${ 6x - 5y = -5 }$

Dividing both sides of the equation by 2, we get:

${ 3x - \frac{5}{2}y = -\frac{5}{2} }$

Now, we can multiply the equation by 2 to make the coefficients of y in both equations equal:

${ 2 \left( 3x - \frac{5}{2}y \right) = 2 \left( -\frac{5}{2} \right) }$

Simplifying the equation, we get:

${ 6x - 5y = -5 }$

Now, we can add the two equations together to eliminate the variable y:

${ (6x - 5y) + (6x - 5y) = (-5) + (-5) }$

Simplifying the equation, we get:

${ 12x - 10y = -10 }$

Now, we can divide both sides of the equation by 2 to simplify the equation:

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables.

Q: What are the methods for solving a system of linear equations?

A: There are two common methods for solving a system of linear equations: the method of substitution and the method of elimination.

Q: What is the method of substitution?

A: The method of substitution involves solving one equation for a variable and then substituting that expression into the other equation.

Q: What is the method of elimination?

A: The method of elimination involves adding or subtracting the equations to eliminate one of the variables.

Q: How do I choose which method to use?

A: The choice of method depends on the specific system of linear equations. If the equations are easy to solve, the method of substitution may be faster. If the equations are more complex, the method of elimination may be more effective.

Q: What if the system of linear equations has no solution?

A: If the system of linear equations has no solution, it means that the equations are inconsistent and there is no value of the variables that can satisfy both equations.

Q: What if the system of linear equations has an infinite number of solutions?

A: If the system of linear equations has an infinite number of solutions, it means that the equations are dependent and there are an infinite number of values of the variables that can satisfy both equations.

Q: How do I know if the system of linear equations has a unique solution?

A: If the system of linear equations has a unique solution, it means that there is only one value of the variables that can satisfy both equations.

Q: Can I use a calculator to solve a system of linear equations?

A: Yes, you can use a calculator to solve a system of linear equations. Many calculators have built-in functions for solving systems of linear equations.

Q: Can I use a computer program to solve a system of linear equations?

A: Yes, you can use a computer program to solve a system of linear equations. Many computer programs, such as MATLAB and Python, have built-in functions for solving systems of linear equations.

Q: What are some common mistakes to avoid when solving a system of linear equations?

A: Some common mistakes to avoid when solving a system of linear equations include:

• Not checking if the equations are consistent before solving them
• Not using the correct method for solving the system of linear equations
• Not checking if the solution is unique before accepting it
• Not checking if the solution satisfies both equations

Q: How do I check if the solution is unique?

A: To check if the solution is unique, you can substitute the solution into both equations and check if it satisfies both equations.

Q: How do I check if the solution satisfies both equations?

A: To check if the solution satisfies both equations, you can substitute the solution into both equations and check if the resulting equations are true.

Q: What if I make a mistake when solving a system of linear equations?

A: If you make a mistake when solving a system of linear equations, you can try to identify the mistake and correct it. If you are unable to correct the mistake, you may need to start over from the beginning.

Q: Can I use a system of linear equations to model real-world problems?

A: Yes, you can use a system of linear equations to model real-world problems. Many real-world problems can be represented as a system of linear equations, and solving the system of linear equations can provide valuable insights into the problem.

Q: What are some examples of real-world problems that can be modeled using a system of linear equations?

A: Some examples of real-world problems that can be modeled using a system of linear equations include:

• Finding the intersection of two lines
• Finding the equation of a line that passes through two points
• Finding the equation of a circle that passes through three points
• Finding the equation of a parabola that passes through four points

Q: How do I know if a system of linear equations is consistent or inconsistent?

A: To determine if a system of linear equations is consistent or inconsistent, you can use the following criteria:

• If the system of linear equations has a unique solution, it is consistent.
• If the system of linear equations has an infinite number of solutions, it is consistent.
• If the system of linear equations has no solution, it is inconsistent.

Q: How do I know if a system of linear equations is dependent or independent?

A: To determine if a system of linear equations is dependent or independent, you can use the following criteria:

• If the system of linear equations has a unique solution, it is independent.
• If the system of linear equations has an infinite number of solutions, it is dependent.

Q: Can I use a system of linear equations to solve a problem that involves multiple variables?

A: Yes, you can use a system of linear equations to solve a problem that involves multiple variables. Many problems that involve multiple variables can be represented as a system of linear equations, and solving the system of linear equations can provide valuable insights into the problem.

Q: What are some common applications of systems of linear equations?

A: Some common applications of systems of linear equations include:

• Finding the intersection of two lines
• Finding the equation of a line that passes through two points
• Finding the equation of a circle that passes through three points
• Finding the equation of a parabola that passes through four points
• Modeling real-world problems that involve multiple variables

Q: Can I use a system of linear equations to solve a problem that involves non-linear equations?

A: No, you cannot use a system of linear equations to solve a problem that involves non-linear equations. Systems of linear equations are used to solve problems that involve linear equations, and non-linear equations require different methods to solve.

Q: What are some common mistakes to avoid when using a system of linear equations to solve a problem?

A: Some common mistakes to avoid when using a system of linear equations to solve a problem include:

• Not checking if the equations are consistent before solving them
• Not using the correct method for solving the system of linear equations
• Not checking if the solution is unique before accepting it
• Not checking if the solution satisfies both equations

Q: How do I know if a system of linear equations is solvable?

A: To determine if a system of linear equations is solvable, you can use the following criteria:

• If the system of linear equations has a unique solution, it is solvable.
• If the system of linear equations has an infinite number of solutions, it is solvable.
• If the system of linear equations has no solution, it is not solvable.

Q: Can I use a system of linear equations to solve a problem that involves multiple equations?

A: Yes, you can use a system of linear equations to solve a problem that involves multiple equations. Many problems that involve multiple equations can be represented as a system of linear equations, and solving the system of linear equations can provide valuable insights into the problem.

Q: What are some common applications of systems of linear equations in science and engineering?

A: Some common applications of systems of linear equations in science and engineering include:

• Modeling population growth
• Modeling chemical reactions
• Modeling electrical circuits
• Modeling mechanical systems
• Modeling thermal systems

Q: Can I use a system of linear equations to solve a problem that involves non-linear equations and multiple variables?

A: No, you cannot use a system of linear equations to solve a problem that involves non-linear equations and multiple variables. Systems of linear equations are used to solve problems that involve linear equations, and non-linear equations and multiple variables require different methods to solve.

Q: What are some common mistakes to avoid when using a system of linear equations to solve a problem in science and engineering?

A: Some common mistakes to avoid when using a system of linear equations to solve a problem in science and engineering include:

• Not checking if the equations are consistent before solving them
• Not using the correct method for solving the system of linear equations
• Not checking if the solution is unique before accepting it
• Not checking if the solution satisfies both equations

Q: How do I know if a system of linear equations is suitable for a problem in science and engineering?

A: To determine if a system of linear equations is suitable for a problem in science and engineering, you can use the following criteria:

• If the problem involves linear equations, a system of linear equations may be suitable.
• If the problem involves non-linear equations, a system of linear equations may not be suitable.
• If the problem involves multiple variables, a system of linear equations may be suitable.

Q: Can I use a system of linear equations to solve a problem that involves real-world data?

A: Yes, you can use a system of linear equations to solve a problem that involves real-world data. Many problems that involve real-world data can be represented as a system of linear equations, and solving the system of linear equations can provide valuable insights into the problem.

Q: What are some common applications of systems of linear equations in economics and finance?

A: Some common applications of systems of linear equations in economics and finance include:

• Modeling economic growth
• Modeling inflation
• Modeling interest rates
• Modeling stock prices
• Modeling