x +y+3z =0 
y-32x+5z = 0
4x-5y+33z= 0

solution of this

We are given the following system of linear equations:

1. \( x + y + 3z = 0 \)
2. \( y - 32x + 5z = 0 \)
3. \( 4x - 5y + 33z = 0 \)

We can solve this system using methods such as substitution, elimination, or matrix methods (like Gaussian elimination). Let's proceed step-by-step using the elimination method.

### Step 1: Solve the first equation for \( y \)
From the first equation, \( x + y + 3z = 0 \), we can solve for \( y \):

\[
y = -x - 3z
\]

### Step 2: Substitute \( y = -x - 3z \) into the second equation
The second equation is \( y - 32x + 5z = 0 \). Substituting \( y = -x - 3z \) into this equation:

\[
(-x - 3z) - 32x + 5z = 0
\]

Simplifying:

\[
-x - 3z - 32x + 5z = 0
\]

\[
-33x + 2z = 0
\]

So, we have:

\[
33x = 2z
\]

\[
x = \frac{2}{33}z
\]

### Step 3: Substitute \( x = \frac{2}{33}z \) and \( y = -x - 3z \) into the third equation
Now, substitute \( x = \frac{2}{33}z \) and \( y = -x - 3z \) into the third equation \( 4x - 5y + 33z = 0 \):

First, calculate \( y \):
\[
y = -\left(\frac{2}{33}z\right) - 3z = -\frac{2}{33}z - 3z = -\frac{2 + 99}{33}z = -\frac{101}{33}z
\]

Now, substitute into the third equation:
\[
4\left(\frac{2}{33}z\right) - 5\left(-\frac{101}{33}z\right) + 33z = 0
\]

Simplifying each term:
\[
\frac{8}{33}z + \frac{505}{33}z + 33z = 0
\]

Now, combine the terms:
\[
\frac{8 + 505}{33}z + 33z = 0
\]
\[
\frac{513}{33}z + 33z = 0
\]

Convert \( 33z \) to a fraction:
\[
\frac{513}{33}z + \frac{1089}{33}z = 0
\]

Now, combine:
\[
\frac{513 + 1089}{33}z = 0
\]
\[
\frac{1602}{33}z = 0
\]

Thus, \( z = 0 \).

### Step 4: Solve for \( x \) and \( y \)
Since \( z = 0 \), substitute \( z = 0 \) into the expressions for \( x \) and \( y \):

\[
x = \frac{2}{33}z = 0
\]
\[
y = -x - 3z = 0
\]

### Final Solution:
The solution to the system of equations is:
\[
x = 0, \quad y = 0, \quad z = 0
\]

Would you like further details on this method, or do you have any questions?

Here are 5 related questions to explore:
1. How does Gaussian elimination work for solving systems of linear equations?
2. What are the conditions for a system of linear equations to have no solution, one solution, or infinitely many solutions?
3. Can matrix methods like Cramer's Rule be used to solve this system? If yes, how?
4. How do linear transformations relate to systems of linear equations?
5. What is the geometric interpretation of a system of three linear equations in three variables?

**Tip**: Always check the determinant of the coefficient matrix when solving a system using matrix methods—it indicates whether the system has a unique solution.

x + y + 3z = 0 
y - 32x + 5z = 0 
4x - 5y + 33z = 0

Learn how to solve the system of linear equations x + y + 3z = 0, y - 32x + 5z = 0, and 4x - 5y + 33z = 0 using the elimination method. This solution covers key concepts in linear algebra, including substitution and elimination methods. Suitable for students in Grades 11-12.

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