Solve the system of equations:
18x + 3y -3z = 0
13x - 2y - 3z = 0

We are given the following system of equations:

\[
\begin{aligned}
18x + 3y - 3z &= 0 \quad \text{(Equation 1)} \\
13x - 2y - 3z &= 0 \quad \text{(Equation 2)} \\
\end{aligned}
\]

Let's solve this system step-by-step.

### Step 1: Express one variable in terms of the others from one equation.
From Equation 1:

\[
18x + 3y - 3z = 0
\]

We can solve for \( y \) in terms of \( x \) and \( z \):

\[
3y = 3z - 18x
\]

\[
y = z - 6x \quad \text{(Equation 3)}
\]

### Step 2: Substitute this expression into Equation 2.

Substituting \( y = z - 6x \) into Equation 2:

\[
13x - 2(z - 6x) - 3z = 0
\]

Distribute the \(-2\) across the terms in the parentheses:

\[
13x - 2z + 12x - 3z = 0
\]

### Step 3: Combine like terms.

\[
25x - 5z = 0
\]

### Step 4: Solve for \( z \) in terms of \( x \).

\[
z = 5x
\]

### Step 5: Substitute \( z = 5x \) back into Equation 3 to solve for \( y \).

\[
y = z - 6x
\]

\[
y = 5x - 6x
\]

\[
y = -x
\]

### Final Answer:
The solution to the system is:

\[
\boxed{(x, y, z) = (x, -x, 5x)}
\]

This represents a family of solutions where \( x \) can be any real number.

---

Would you like more details or have any questions?

### Related Questions:
1. How can I solve systems of three equations with three variables?
2. What is the geometric interpretation of this system of equations?
3. How do you determine if a system of equations has no solution, one solution, or infinitely many solutions?
4. Can substitution always be used to solve linear systems?
5. How does this solution change if the equations are nonlinear?

### Tip:
When solving systems of equations, checking your solution by substituting back into the original equations is a good habit to catch any mistakes.

Learn how to solve the system of equations 18x + 3y - 3z = 0 and 13x - 2y - 3z = 0 with detailed step-by-step solution using the substitution method. Understand the method to express variables and solve for x, y, and z in terms of each other. Suitable for high school students in Grades 10-12.

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