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We are given two simultaneous equations:

1. \(\frac{y - 28}{3} = 6x\)
2. \(2x + y = 8\)

We will solve them using substitution.

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

Start by multiplying both sides of the first equation by 3 to eliminate the fraction:
\[
y - 28 = 18x
\]
Now, solve for \(y\):
\[
y = 18x + 28
\]

### Step 2: Substitute this expression for \(y\) into the second equation

Now that we have \(y = 18x + 28\), substitute this into the second equation \(2x + y = 8\):
\[
2x + (18x + 28) = 8
\]
Simplify the equation:
\[
2x + 18x + 28 = 8
\]
\[
20x + 28 = 8
\]
Subtract 28 from both sides:
\[
20x = 8 - 28
\]
\[
20x = -20
\]
Now, solve for \(x\):
\[
x = \frac{-20}{20} = -1
\]

### Step 3: Substitute \(x = -1\) into the expression for \(y\)

Now that we know \(x = -1\), substitute it back into the expression \(y = 18x + 28\):
\[
y = 18(-1) + 28 = -18 + 28 = 10
\]

### Final Solution:
The solution to the system is:
\[
x = -1, \quad y = 10
\]

Would you like more details on any of these steps or further clarifications?

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Here are five follow-up questions for you:
1. How do you solve simultaneous equations using elimination instead of substitution?
2. What happens when there is no solution or infinitely many solutions to a system of equations?
3. Can you use the graphing method to verify this solution?
4. How do you identify whether a system of equations is consistent or inconsistent?
5. What is the difference between linear and non-linear simultaneous equations?

**Tip:** When solving simultaneous equations, always double-check your substitution or elimination to avoid simple arithmetic errors.

Solve the simultaneous equations below using substitution:

1. (y - 28) / 3 = 6x
2. 2x + y = 8

Learn how to solve simultaneous equations using the substitution method with the example (y - 28) / 3 = 6x and 2x + y = 8. This step-by-step guide covers key algebraic concepts and is suitable for students in Grades 8-10.

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