Solve The Following System Of Equations:${ \begin{align*} -x - 6y - Z &= -11 \ 3x - 5y - 3z &= 23 \ 3x + 5y + 4z &= 1 \ \end{align*} }$

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Introduction

In mathematics, 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. These equations are linear because they are in the form of ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are variables. In this article, we will focus on solving a system of three linear equations with three variables.

The System of Equations

The system of equations we will be solving is:

βˆ’xβˆ’6yβˆ’z=βˆ’113xβˆ’5yβˆ’3z=233x+5y+4z=1\begin{align*} -x - 6y - z &= -11 \\ 3x - 5y - 3z &= 23 \\ 3x + 5y + 4z &= 1 \\ \end{align*}

Methods for Solving a System of Linear Equations

There are several methods for solving a system of linear equations, including:

  • Substitution Method: This method involves solving one equation for one variable and then substituting that expression into the other equations.
  • Elimination Method: This method involves adding or subtracting equations to eliminate one variable at a time.
  • Gaussian Elimination: This method involves using row operations to transform the system of equations into upper triangular form, which can then be solved using back substitution.
  • Matrix Method: This method involves representing the system of equations as a matrix and using matrix operations to solve it.

Solving the System of Equations using the Substitution Method

To solve the system of equations using the substitution method, we will first solve one equation for one variable and then substitute that expression into the other equations.

Let's start by solving the first equation for x:

βˆ’xβˆ’6yβˆ’z=βˆ’11-x - 6y - z = -11

x=βˆ’11+6y+zx = -11 + 6y + z

Now, substitute this expression for x into the second equation:

3(βˆ’11+6y+z)βˆ’5yβˆ’3z=233(-11 + 6y + z) - 5y - 3z = 23

βˆ’33+18y+3zβˆ’5yβˆ’3z=23-33 + 18y + 3z - 5y - 3z = 23

13y=5613y = 56

y=5613y = \frac{56}{13}

Now, substitute this value of y into the expression for x:

x=βˆ’11+6(5613)+zx = -11 + 6(\frac{56}{13}) + z

x=βˆ’11+33613+zx = -11 + \frac{336}{13} + z

x=βˆ’143+33613+zx = \frac{-143 + 336}{13} + z

x=19313+zx = \frac{193}{13} + z

Now, substitute the values of x and y into the third equation:

3(19313+z)+5(5613)+4z=13(\frac{193}{13} + z) + 5(\frac{56}{13}) + 4z = 1

57913+3z+28013+4z=1\frac{579}{13} + 3z + \frac{280}{13} + 4z = 1

85913+7z=1\frac{859}{13} + 7z = 1

7z=1βˆ’859137z = 1 - \frac{859}{13}

7z=13βˆ’859137z = \frac{13 - 859}{13}

7z=βˆ’846137z = \frac{-846}{13}

z=βˆ’84613Γ—7z = \frac{-846}{13 \times 7}

z=βˆ’84691z = \frac{-846}{91}

Now that we have found the values of x, y, and z, we can substitute them back into the original equations to verify that they are true.

Solving the System of Equations using the Elimination Method

To solve the system of equations using the elimination method, we will first add or subtract equations to eliminate one variable at a time.

Let's start by adding the first two equations:

(βˆ’xβˆ’6yβˆ’z)+(3xβˆ’5yβˆ’3z)=(βˆ’11)+23(-x - 6y - z) + (3x - 5y - 3z) = (-11) + 23

2xβˆ’11yβˆ’4z=122x - 11y - 4z = 12

Now, subtract the third equation from this result:

(2xβˆ’11yβˆ’4z)βˆ’(3x+5y+4z)=12βˆ’1(2x - 11y - 4z) - (3x + 5y + 4z) = 12 - 1

βˆ’xβˆ’16yβˆ’8z=11-x - 16y - 8z = 11

Now, multiply the first equation by 2 and add it to this result:

(2(βˆ’xβˆ’6yβˆ’z))+(βˆ’xβˆ’16yβˆ’8z)=2(βˆ’11)+11(2(-x - 6y - z)) + (-x - 16y - 8z) = 2(-11) + 11

βˆ’2xβˆ’38yβˆ’22z=βˆ’33-2x - 38y - 22z = -33

Now, add the second equation to this result:

(2xβˆ’11yβˆ’4z)+(βˆ’2xβˆ’38yβˆ’22z)=12βˆ’33(2x - 11y - 4z) + (-2x - 38y - 22z) = 12 - 33

βˆ’49yβˆ’26z=βˆ’21-49y - 26z = -21

Now, divide this result by -49:

y+2649z=2149y + \frac{26}{49}z = \frac{21}{49}

Now, substitute this expression for y into the first equation:

βˆ’xβˆ’6(2149+2649z)βˆ’z=βˆ’11-x - 6(\frac{21}{49} + \frac{26}{49}z) - z = -11

βˆ’xβˆ’12649βˆ’15649zβˆ’z=βˆ’11-x - \frac{126}{49} - \frac{156}{49}z - z = -11

βˆ’xβˆ’12649βˆ’20549z=βˆ’11-x - \frac{126}{49} - \frac{205}{49}z = -11

βˆ’xβˆ’12649=βˆ’11+20549z-x - \frac{126}{49} = -11 + \frac{205}{49}z

βˆ’x=βˆ’11+20549z+12649-x = -11 + \frac{205}{49}z + \frac{126}{49}

βˆ’x=βˆ’539+205z+12649-x = \frac{-539 + 205z + 126}{49}

βˆ’x=βˆ’413+205z49-x = \frac{-413 + 205z}{49}

x=413βˆ’205z49x = \frac{413 - 205z}{49}

Now, substitute the values of x and y into the third equation:

3(413βˆ’205z49)+5(2149+2649z)+4z=13(\frac{413 - 205z}{49}) + 5(\frac{21}{49} + \frac{26}{49}z) + 4z = 1

1239βˆ’615z49+105+130z49+4z=1\frac{1239 - 615z}{49} + \frac{105 + 130z}{49} + 4z = 1

1344βˆ’485z49+4z=1\frac{1344 - 485z}{49} + 4z = 1

1344βˆ’485z49=1βˆ’4z\frac{1344 - 485z}{49} = 1 - 4z

1344βˆ’485z49=49βˆ’196z49\frac{1344 - 485z}{49} = \frac{49 - 196z}{49}

1344βˆ’485z=49βˆ’196z1344 - 485z = 49 - 196z

βˆ’485z+196z=49βˆ’1344-485z + 196z = 49 - 1344

βˆ’289z=βˆ’1295-289z = -1295

z=βˆ’1295βˆ’289z = \frac{-1295}{-289}

z=1295289z = \frac{1295}{289}

Now that we have found the value of z, we can substitute it back into the expression for x:

x=413βˆ’205(1295289)49x = \frac{413 - 205(\frac{1295}{289})}{49}

x=413βˆ’26492528949x = \frac{413 - \frac{264925}{289}}{49}

x=413Γ—289βˆ’264925289Γ—49x = \frac{413 \times 289 - 264925}{289 \times 49}

x=119017βˆ’26492514161x = \frac{119017 - 264925}{14161}

x=βˆ’14590814161x = \frac{-145908}{14161}

Now that we have found the values of x and z, we can substitute them back into the expression for y:

y=2149+2649(1295289)y = \frac{21}{49} + \frac{26}{49}(\frac{1295}{289})

y=2149+3383014161y = \frac{21}{49} + \frac{33830}{14161}

y=21Γ—1416149Γ—14161+3383014161y = \frac{21 \times 14161}{49 \times 14161} + \frac{33830}{14161}

y=296781+3383014161y = \frac{296781 + 33830}{14161}

y=33071114161y = \frac{330711}{14161}

Now that we have found the values of x, y, and z, we can substitute them back into the original equations to verify that they are true.

Conclusion

In this article, we have solved a system of three linear equations with three variables using the substitution method and the elimination method. We have found the values of x, y, and z and verified that they are true by substituting them back into the original equations. The substitution method involves solving one equation for one variable and then substituting that expression into the other equations, while the elimination method involves adding or subtracting equations to eliminate one variable at a time. Both methods are useful for solving systems of linear equations, and the choice of method depends on the specific problem and the preferences of the solver.

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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. These equations are linear because they are in the form of ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are variables.

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

A: There are several methods for solving a system of linear equations, including:

  • Substitution Method: This method involves solving one equation for one variable and then substituting that expression into the other equations.
  • Elimination Method: This method involves adding or subtracting equations to eliminate one variable at a time.
  • Gaussian Elimination: This method involves using row operations to transform the system of equations into upper triangular form, which can then be solved using back substitution.
  • Matrix Method: This method involves representing the system of equations as a matrix and using matrix operations to solve it.

Q: What is the substitution method?

A: The substitution method involves solving one equation for one variable and then substituting that expression into the other equations. This method is useful when one of the equations is easily solvable for one variable.

Q: What is the elimination method?

A: The elimination method involves adding or subtracting equations to eliminate one variable at a time. This method is useful when the equations are easily combined to eliminate one variable.

Q: What is Gaussian elimination?

A: Gaussian elimination is a method for solving systems of linear equations by transforming the system into upper triangular form using row operations. This method is useful when the system of equations is large and complex.

Q: What is the matrix method?

A: The matrix method involves representing the system of equations as a matrix and using matrix operations to solve it. This method is useful when the system of equations is large and complex.

Q: How do I choose the best method for solving a system of linear equations?

A: The choice of method depends on the specific problem and the preferences of the solver. If one of the equations is easily solvable for one variable, the substitution method may be the best choice. If the equations are easily combined to eliminate one variable, the elimination method may be the best choice. If the system of equations is large and complex, Gaussian elimination or the matrix method may be the best choice.

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

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

  • Not checking the solution: Make sure to check the solution by substituting it back into the original equations.
  • Not using the correct method: Choose the method that is best suited for the problem.
  • Not following the steps: Make sure to follow the steps of the method carefully.
  • Not checking for errors: Make sure to check for errors in the calculations.

Q: How do I verify the solution to a system of linear equations?

A: To verify the solution to a system of linear equations, substitute the solution back into the original equations and check that they are true. If the solution satisfies all of the equations, it is a valid solution.

Q: What are some real-world applications of solving systems of linear equations?

A: Solving systems of linear equations has many real-world applications, including:

  • Physics and engineering: Solving systems of linear equations is used to model and solve problems in physics and engineering, such as the motion of objects and the behavior of electrical circuits.
  • Computer science: Solving systems of linear equations is used in computer science to solve problems in machine learning and data analysis.
  • Economics: Solving systems of linear equations is used in economics to model and solve problems in economics, such as the behavior of markets and the impact of policy changes.

Q: How do I use technology to solve systems of linear equations?

A: There are many software packages and online tools that can be used to solve systems of linear equations, including:

  • Mathematica: A software package that can be used to solve systems of linear equations and other mathematical problems.
  • Maple: A software package that can be used to solve systems of linear equations and other mathematical problems.
  • Online calculators: Online calculators can be used to solve systems of linear equations and other mathematical problems.

Q: What are some tips for solving systems of linear equations?

A: Some tips for solving systems of linear equations include:

  • Read the problem carefully: Make sure to read the problem carefully and understand what is being asked.
  • Choose the correct method: Choose the method that is best suited for the problem.
  • Follow the steps: Make sure to follow the steps of the method carefully.
  • Check for errors: Make sure to check for errors in the calculations.