Type The Correct Answer In Each Box.Find The Solution To This System Of Equations.$\[ \begin{array}{l} 3x + 2y + 3z = 3 \\ 4x - 5y + 7z = 1 \\ 2x + 3y - 2z = 6 \end{array} \\]$\[ x = \square \\]$\[ y = \square \\]$\[ z =

Introduction

Solving a system of linear equations is a fundamental concept in mathematics, particularly in algebra and linear algebra. It involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we will explore how to solve a system of three linear equations with three variables using the method of substitution and elimination.

The System of Equations

The given system of equations is:

{ \begin{array}{l} 3x + 2y + 3z = 3 \\ 4x - 5y + 7z = 1 \\ 2x + 3y - 2z = 6 \end{array} \}

We are asked to find the values of $x$, $y$, and $z$ that satisfy all three equations simultaneously.

Method of Substitution

One way to solve this system of equations is by using the method of substitution. This 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 $x$:

$3x + 2y + 3z = 3$

Subtracting $2y + 3z$ from both sides gives:

$3x = 3 - 2y - 3z$

Dividing both sides by 3 gives:

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

Now, we can substitute this expression for $x$ into the second and third equations:

{ \begin{array}{l} 4\left(\frac{3 - 2y - 3z}{3}\right) - 5y + 7z = 1 \\ 2\left(\frac{3 - 2y - 3z}{3}\right) + 3y - 2z = 6 \end{array} \}

Simplifying these equations gives:

{ \begin{array}{l} \frac{4(3 - 2y - 3z)}{3} - 5y + 7z = 1 \\ \frac{2(3 - 2y - 3z)}{3} + 3y - 2z = 6 \end{array} \}

Multiplying both sides of each equation by 3 to eliminate the fractions gives:

{ \begin{array}{l} 4(3 - 2y - 3z) - 15y + 21z = 3 \\ 2(3 - 2y - 3z) + 9y - 6z = 18 \end{array} \}

Expanding and simplifying these equations gives:

{ \begin{array}{l} 12 - 8y - 12z - 15y + 21z = 3 \\ 6 - 4y - 6z + 9y - 6z = 18 \end{array} \}

Combining like terms gives:

{ \begin{array}{l} -23y + 9z = -9 \\ 5y - 12z = 12 \end{array} \}

Method of Elimination

Another way to solve this system of equations is by using the method of elimination. This involves multiplying the equations by necessary multiples such that the coefficients of one variable (either $y$ or $z$) are the same in both equations.

Let's multiply the first equation by 5 and the second equation by 23:

{ \begin{array}{l} -115y + 45z = -45 \\ 115y - 276z = 276 \end{array} \}

Now, we can add both equations to eliminate the variable $y$:

$-231z = 231$

Dividing both sides by -231 gives:

$z = -1$

Now that we have found the value of $z$, we can substitute it into one of the original equations to find the value of $y$. Let's use the second equation:

$5y - 12z = 12$

Substituting $z = -1$ gives:

$5y + 12 = 12$

Subtracting 12 from both sides gives:

$5y = 0$

Dividing both sides by 5 gives:

$y = 0$

Now that we have found the values of $y$ and $z$, we can substitute them into one of the original equations to find the value of $x$. Let's use the first equation:

$3x + 2y + 3z = 3$

Substituting $y = 0$ and $z = -1$ gives:

$3x + 2(0) + 3(-1) = 3$

Simplifying gives:

$3x - 3 = 3$

Adding 3 to both sides gives:

$3x = 6$

Dividing both sides by 3 gives:

$x = 2$

Conclusion

In this article, we have solved a system of three linear equations with three variables using the method of substitution and elimination. We have found the values of $x$, $y$, and $z$ that satisfy all three equations simultaneously. The values are $x = 2$, $y = 0$, and $z = -1$.

Discussion

Solving a system of linear equations is an important concept in mathematics, particularly in algebra and linear algebra. It involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we have used the method of substitution and elimination to solve a system of three linear equations with three variables. The method of substitution involves solving one equation for one variable and then substituting that expression into the other equations. The method of elimination involves multiplying the equations by necessary multiples such that the coefficients of one variable are the same in both equations.

Applications

Solving a system of linear equations has many applications in real-life situations. For example, it can be used to find the values of unknowns in a system of equations that represent a physical situation. It can also be used to find the values of unknowns in a system of equations that represent a financial situation.

Future Work

In the future, we can explore other methods of solving systems of linear equations, such as the method of matrices and the method of determinants. We can also explore the applications of solving systems of linear equations in real-life situations, such as in physics, engineering, and finance.

References

• [1] "Linear Algebra and Its Applications" by Gilbert Strang
• [2] "Introduction to Linear Algebra" by Jim Hefferon
• [3] "Linear Algebra: A Modern Introduction" by David Poole
  

Introduction

Solving a system of linear equations is a fundamental concept in mathematics, particularly in algebra and linear algebra. In our previous article, we explored how to solve a system of three linear equations with three variables using the method of substitution and elimination. In this article, we will answer some frequently asked questions about 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 two or more variables. Each equation is a linear equation, which means that it can be written in the form ax + by + cz = d, where a, b, c, and d are constants, and x, y, and z are variables.

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 some values of the variables, then the system has a solution. If the equations are not true for any values of the variables, then the system has no solution.

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

A: A consistent system of linear equations is one that has a solution. An inconsistent system of linear equations is one that has no solution.

Q: How do I solve a system of linear equations using the method of substitution?

A: To solve a system of linear equations using the method of substitution, you need to solve one equation for one variable and then substitute that expression into the other equations. This will give you a new system of equations with one less variable.

Q: How do I solve a system of linear equations using the method of elimination?

A: To solve a system of linear equations using the method of elimination, you need to multiply the equations by necessary multiples such that the coefficients of one variable are the same in both equations. Then, you can add or subtract the equations to eliminate the variable.

Q: What is the difference between the method of substitution and the method of elimination?

A: The method of substitution involves solving one equation for one variable and then substituting that expression into the other equations. The method of elimination involves multiplying the equations by necessary multiples such that the coefficients of one variable are the same in both equations.

Q: Can I use both the method of substitution and the method of elimination to solve a system of linear equations?

A: Yes, you can use both the method of substitution and the method of elimination to solve a system of linear equations. In fact, using both methods can help you to check your work and ensure that you have found the correct solution.

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 trying to solve them
• Not following the correct order of operations when solving the equations
• Not checking if the solution satisfies all of the equations
• Not using the correct method of substitution or elimination

Q: How do I know if I have found the correct solution to a system of linear equations?

A: To know if you have found the correct solution to a system of linear equations, you need to check if the solution satisfies all of the equations. You can do this by plugging the solution back into each of the equations and checking if they are true.

Conclusion

Solving a system of linear equations is an important concept in mathematics, particularly in algebra and linear algebra. In this article, we have answered some frequently asked questions about solving systems of linear equations. We hope that this article has been helpful in clarifying any confusion you may have had about solving systems of linear equations.

Discussion

Solving a system of linear equations is a fundamental concept in mathematics, and it has many applications in real-life situations. In this article, we have explored some of the common mistakes to avoid when solving a system of linear equations and how to check if you have found the correct solution.

Applications

Solving a system of linear equations has many applications in real-life situations, such as in physics, engineering, and finance. It can be used to find the values of unknowns in a system of equations that represent a physical situation, or to find the values of unknowns in a system of equations that represent a financial situation.

Future Work

In the future, we can explore other methods of solving systems of linear equations, such as the method of matrices and the method of determinants. We can also explore the applications of solving systems of linear equations in real-life situations, such as in physics, engineering, and finance.

References

• [1] "Linear Algebra and Its Applications" by Gilbert Strang
• [2] "Introduction to Linear Algebra" by Jim Hefferon
• [3] "Linear Algebra: A Modern Introduction" by David Poole