We are given the system of equations:

\[
\begin{aligned}
1) &\quad 4x - 6y = 5 \\
2) &\quad x - 3y = 6
\end{aligned}
\]

To solve for \( y \), follow these steps:

### Step 1: Solve for \( x \) in terms of \( y \) from equation (2)

From equation (2), we can solve for \( x \) as follows:

\[
x = 6 + 3y
\]

### Step 2: Substitute \( x = 6 + 3y \) into equation (1)

Now, substitute \( x = 6 + 3y \) into equation (1):

\[
4(6 + 3y) - 6y = 5
\]

### Step 3: Simplify the equation

Expand the terms:

\[
24 + 12y - 6y = 5
\]

Combine like terms:

\[
24 + 6y = 5
\]

### Step 4: Solve for \( y \)

Isolate \( y \):

\[
6y = 5 - 24
\]

\[
6y = -19
\]

\[
y = \frac{-19}{6}
\]

So, the solution for \( y \) is:

\[
y = \frac{-19}{6}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do we solve for \( x \) once \( y \) is known?
2. What is the general method for solving systems of linear equations?
3. Can this system be solved using substitution or elimination? Which is better?
4. How do you verify the solution to a system of linear equations?
5. What happens if a system of equations has no solution or infinite solutions?

**Tip:** Always check your solution by substituting it back into the original equations!

Learn how to solve a system of linear equations using substitution. This step-by-step guide demonstrates how to find the value of y in the equations 4x - 6y = 5 and x - 3y = 6. Ideal for students in Grades 8-10 learning algebra and systems of equations.

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