x
​

−
−
​

y
y
​

−
+
+
​

6z
4z
5z
​

=2
=0
=0
​

It looks like you've provided a system of three linear equations. Here they are written out:

\[
\begin{aligned}
x - y + 6z &= 2, \\
- y + 4z &= 0, \\
y + 5z &= 0.
\end{aligned}
\]

Let's solve this system step by step.

### Step 1: Solve the second equation for \( y \).

\[
- y + 4z = 0 \implies y = 4z
\]

### Step 2: Substitute \( y = 4z \) into the third equation.

\[
4z + 5z = 0 \implies 9z = 0 \implies z = 0
\]

### Step 3: Substitute \( z = 0 \) back into the expression for \( y \).

\[
y = 4z = 4(0) = 0
\]

### Step 4: Substitute \( z = 0 \) and \( y = 0 \) into the first equation to solve for \( x \).

\[
x - 0 + 6(0) = 2 \implies x = 2
\]

### Solution:
\[
\boxed{x = 2}, \quad \boxed{y = 0}, \quad \boxed{z = 0}
\]

This is the unique solution to the system of equations.

Would you like any more details or have any questions?

Here are 8 related questions you might find interesting:
1. What if \( z \) was not zero—how would that change the solution?
2. How can these equations be solved using matrices?
3. What is the geometric interpretation of this system of equations?
4. Can you express this system in vector form?
5. What if the equations were dependent—how would that affect the solution?
6. How do you determine if a system of equations has no solution, one solution, or infinitely many solutions?
7. How does Gaussian elimination solve this system of equations?
8. How can substitution and elimination methods be compared?

**Tip:** Always double-check your substitutions when solving systems of equations to avoid errors.

Explore how to solve a system of three linear equations step-by-step using the substitution method. This problem involves fundamental concepts of algebra, including linear equations and substitution. Suitable for high school students in Grades 9-12.

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