Solve The System Of Equations:$\[ \begin{array}{l} 7x + 12y + Z = 96 \\ -4x - 6y - Z = -57 \\ 9x + 15y + Z = 120 \\ \end{array} \\]Find The Values Of:$\[ x = \\ y = \\ z = \\]

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Introduction

In mathematics, a system of linear equations is a set of two or more linear equations that involve the same set of variables. Solving a system of linear equations involves finding the values of the variables that satisfy all the equations in the system. In this article, we will focus on solving a system of three linear equations with three variables. We will use the method of substitution and elimination to find the values of the variables.

The System of Equations

The system of equations we will be solving is:

{ \begin{array}{l} 7x + 12y + z = 96 \\ -4x - 6y - z = -57 \\ 9x + 15y + z = 120 \\ \end{array} \}

This system of equations has three variables: x, y, and z. We need to find the values of these variables that satisfy all three equations.

Method of Substitution

One way to solve this system of equations is to use the method of substitution. This method involves solving one equation for one variable and then substituting that expression into the other equations.

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

7x+12y+z=96{ 7x + 12y + z = 96 }

Subtracting 7x and 12y from both sides gives:

z=96βˆ’7xβˆ’12y{ z = 96 - 7x - 12y }

Now, we can substitute this expression for z into the other two equations.

Substituting into the Second Equation

Substituting the expression for z into the second equation gives:

βˆ’4xβˆ’6yβˆ’(96βˆ’7xβˆ’12y)=βˆ’57{ -4x - 6y - (96 - 7x - 12y) = -57 }

Expanding and simplifying this equation gives:

βˆ’4xβˆ’6yβˆ’96+7x+12y=βˆ’57{ -4x - 6y - 96 + 7x + 12y = -57 }

Combine like terms:

3x+6yβˆ’96=βˆ’57{ 3x + 6y - 96 = -57 }

Add 96 to both sides:

3x+6y=39{ 3x + 6y = 39 }

Substituting into the Third Equation

Substituting the expression for z into the third equation gives:

9x+15y+(96βˆ’7xβˆ’12y)=120{ 9x + 15y + (96 - 7x - 12y) = 120 }

Expanding and simplifying this equation gives:

9x+15y+96βˆ’7xβˆ’12y=120{ 9x + 15y + 96 - 7x - 12y = 120 }

Combine like terms:

2x+3y+96=120{ 2x + 3y + 96 = 120 }

Subtract 96 from both sides:

2x+3y=24{ 2x + 3y = 24 }

Solving the System of Equations

Now we have two equations with two variables:

3x+6y=39{ 3x + 6y = 39 }

2x+3y=24{ 2x + 3y = 24 }

We can solve this system of equations using the method of substitution or elimination. Let's use the method of elimination.

Elimination Method

To eliminate one variable, we need to multiply one or both of the equations by a number that will make the coefficients of one variable the same.

Let's multiply the second equation by 2:

4x+6y=48{ 4x + 6y = 48 }

Now, we can subtract the first equation from this new equation:

(4x+6y)βˆ’(3x+6y)=48βˆ’39{ (4x + 6y) - (3x + 6y) = 48 - 39 }

This simplifies to:

x=9{ x = 9 }

Finding the Values of y and z

Now that we have the value of x, we can substitute it into one of the original equations to find the value of y.

Let's substitute x = 9 into the first equation:

7(9)+12y+z=96{ 7(9) + 12y + z = 96 }

Expanding and simplifying this equation gives:

63+12y+z=96{ 63 + 12y + z = 96 }

Subtracting 63 from both sides gives:

12y+z=33{ 12y + z = 33 }

Now, we can substitute x = 9 into the second equation:

βˆ’4(9)βˆ’6yβˆ’z=βˆ’57{ -4(9) - 6y - z = -57 }

Expanding and simplifying this equation gives:

βˆ’36βˆ’6yβˆ’z=βˆ’57{ -36 - 6y - z = -57 }

Adding 36 to both sides gives:

βˆ’6yβˆ’z=βˆ’21{ -6y - z = -21 }

Solving for y and z

Now we have two equations with two variables:

12y+z=33{ 12y + z = 33 }

βˆ’6yβˆ’z=βˆ’21{ -6y - z = -21 }

We can solve this system of equations using the method of substitution or elimination. Let's use the method of addition.

Adding the two equations gives:

(12y+z)+(βˆ’6yβˆ’z)=33+(βˆ’21){ (12y + z) + (-6y - z) = 33 + (-21) }

This simplifies to:

6y=12{ 6y = 12 }

Dividing both sides by 6 gives:

y=2{ y = 2 }

Finding the Value of z

Now that we have the value of y, we can substitute it into one of the equations to find the value of z.

Let's substitute y = 2 into the equation:

12y+z=33{ 12y + z = 33 }

Expanding and simplifying this equation gives:

12(2)+z=33{ 12(2) + z = 33 }

This simplifies to:

24+z=33{ 24 + z = 33 }

Subtracting 24 from both sides gives:

z=9{ z = 9 }

Conclusion

In this article, we solved a system of three linear equations with three variables using the method of substitution and elimination. We found the values of x, y, and z to be x = 9, y = 2, and z = 9.

Final Answer

The final answer is:

x=9{ x = 9 }

y=2{ y = 2 }

z=9{ z = 9 }

Discussion

This system of equations can be solved using other methods such as the method of matrices or the method of Cramer's rule. The method of substitution and elimination is a simple and straightforward method that can be used to solve systems of linear equations.

Applications

Systems of linear equations have many applications in mathematics and science. They are used to model real-world problems such as the motion of objects, the flow of fluids, and the behavior of electrical circuits.

Future Work

In the future, we can explore other methods of solving systems of linear equations such as the method of matrices or the method of Cramer's rule. We can also apply these methods to more complex systems of equations and explore their applications in mathematics and science.

Introduction

In our previous article, we solved a system of three linear equations with three variables using the method of substitution and elimination. In this article, we will answer some common questions that students often have when solving systems of linear equations.

Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that involve the same set of variables. For example:

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

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

A: A system of linear equations has a solution if and only if the equations are consistent. In other words, if the equations are true for the same values of the variables, then the system has a solution.

Q: What is the difference between a system of linear equations and a system of nonlinear equations?

A: A system of linear equations is a set of linear equations, whereas a system of nonlinear equations is a set of nonlinear equations. For example:

{ \begin{array}{l} x^2 + y^2 = 4 \\ x + y = 2 \\ \end{array} \}

This is a system of nonlinear equations because the equations involve powers of the variables.

Q: How do I solve a system of linear equations?

A: There are several methods to solve a system of linear equations, including the method of substitution, the method of elimination, and the method of matrices. The method of substitution involves solving one equation for one variable and then substituting that expression into the other equations.

Q: What is the method of elimination?

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

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

We can add the two equations to eliminate the variable y:

(2x+3y)+(xβˆ’2y)=7+(βˆ’3){ (2x + 3y) + (x - 2y) = 7 + (-3) }

This simplifies to:

3x=4{ 3x = 4 }

Q: What is the method of matrices?

A: The method of matrices involves representing the system of linear equations as a matrix and then using row operations to solve the system. For example:

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

We can represent the system as a matrix:

[231βˆ’2][xy]=[7βˆ’3]{ \begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -3 \end{bmatrix} }

Q: What is the method of Cramer's rule?

A: The method of Cramer's rule involves using determinants to solve the system of linear equations. For example:

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

We can use Cramer's rule to find the values of x and y.

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 if the equations are consistent before solving the system
  • Not using the correct method to solve the system
  • Not checking if the solution satisfies all the equations
  • Not simplifying the equations before solving the system

Conclusion

In this article, we answered some common questions that students often have when solving systems of linear equations. We also discussed some common mistakes to avoid when solving systems of linear equations.

Final Answer

The final answer is:

  • A system of linear equations is a set of two or more linear equations that involve the same set of variables.
  • A system of linear equations has a solution if and only if the equations are consistent.
  • The method of substitution, the method of elimination, and the method of matrices are some common methods to solve a system of linear equations.
  • Cramer's rule is a method to solve a system of linear equations using determinants.
  • Some common mistakes to avoid when solving systems of linear equations include not checking if the equations are consistent, not using the correct method, not checking if the solution satisfies all the equations, and not simplifying the equations before solving the system.

Discussion

Systems of linear equations have many applications in mathematics and science. They are used to model real-world problems such as the motion of objects, the flow of fluids, and the behavior of electrical circuits.

Future Work

In the future, we can explore other methods of solving systems of linear equations such as the method of matrices or the method of Cramer's rule. We can also apply these methods to more complex systems of equations and explore their applications in mathematics and science.