Let's solve each equation step by step.

### Equation 1: \( \frac{1}{4}(x - 28) = 2x \)

1. **Distribute** the \( \frac{1}{4} \) to both terms inside the parentheses:
   \[
   \frac{1}{4}x - \frac{1}{4} \times 28 = 2x
   \]
   \[
   \frac{1}{4}x - 7 = 2x
   \]

2. **Move** the \( \frac{1}{4}x \) term to the right side by subtracting \( \frac{1}{4}x \) from both sides:
   \[
   -7 = 2x - \frac{1}{4}x
   \]

3. **Combine** the like terms on the right side:
   \[
   -7 = \left(2 - \frac{1}{4}\right)x
   \]
   \[
   -7 = \frac{8}{4}x - \frac{1}{4}x
   \]
   \[
   -7 = \frac{7}{4}x
   \]

4. **Solve** for \( x \) by multiplying both sides by the reciprocal of \( \frac{7}{4} \) (which is \( \frac{4}{7} \)):
   \[
   x = -7 \times \frac{4}{7}
   \]
   \[
   x = -4
   \]

### Equation 2: \( 13 - 5x = -7.5x - t \)

1. **Move** the \( -7.5x \) term to the left side by adding \( 7.5x \) to both sides:
   \[
   13 - 5x + 7.5x = -t
   \]
   \[
   13 + 2.5x = -t
   \]

2. **Isolate** \( t \) by moving the terms around:
   \[
   t = -13 - 2.5x
   \]

### Final Solutions:
- \( x = -4 \) from the first equation.
- \( t = -13 - 2.5x \) from the second equation.

You can substitute \( x = -4 \) into the second equation to find the value of \( t \):

\[
t = -13 - 2.5(-4)
\]
\[
t = -13 + 10
\]
\[
t = -3
\]

So the final answers are:
- \( x = -4 \)
- \( t = -3 \)

Would you like more details on any of these steps? 

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**Relative Questions:**
1. What happens if you change the constant in the first equation?
2. How do you solve equations with variables on both sides?
3. Can you solve for \( x \) if the second equation had no \( t \) term?
4. What are the implications if \( t \) were a positive number in the second equation?
5. How do you interpret the solution graphically?

**Tip:** Always check your solutions by substituting them back into the original equations.

Learn how to solve the equations 1/4(x-28) = 2x and 13 - 5x = -7.5x - t step-by-step. This problem covers linear equations and involves the distributive property. Suitable for Grades 7-9 students.

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