We are given two equations:

1. \( 2x + 3y = 5p \)
2. \( y = 2x + p \)

Let's solve this system of equations for \( x \) and \( y \) in terms of \( p \).

### Step 1: Substitute the expression for \( y \) from the second equation into the first equation.

Substitute \( y = 2x + p \) into \( 2x + 3y = 5p \):

\[
2x + 3(2x + p) = 5p
\]

### Step 2: Expand the equation.

Distribute the 3 on the left-hand side:

\[
2x + 6x + 3p = 5p
\]

Combine like terms:

\[
8x + 3p = 5p
\]

### Step 3: Solve for \( x \).

Subtract \( 3p \) from both sides:

\[
8x = 5p - 3p
\]

\[
8x = 2p
\]

Now divide both sides by 8:

\[
x = \frac{2p}{8} = \frac{p}{4}
\]

### Step 4: Substitute \( x = \frac{p}{4} \) into the second equation.

Substitute \( x = \frac{p}{4} \) into \( y = 2x + p \):

\[
y = 2\left(\frac{p}{4}\right) + p
\]

Simplify:

\[
y = \frac{2p}{4} + p = \frac{p}{2} + p
\]

\[
y = \frac{p}{2} + \frac{2p}{2} = \frac{3p}{2}
\]

### Final Solution:

The solutions for \( x \) and \( y \) in terms of \( p \) are:

\[
x = \frac{p}{4}, \quad y = \frac{3p}{2}
\]

Would you like more details or have any further questions?

### Related Questions:

1. How can this solution be interpreted geometrically?
2. What happens if \( p = 0 \) in this system of equations?
3. How would the solution change if \( y = 3x + p \) instead?
4. Can you derive the values of \( x \) and \( y \) if a specific value for \( p \) is provided, say \( p = 2 \)?
5. How would you solve this system using matrix methods?

### Tip:
When solving systems of linear equations, substitution is a powerful method when one equation is already solved for a variable.

Learn how to solve a system of linear equations using the substitution method. This problem involves solving 2x + 3y = 5p and y = 2x + p step-by-step, suitable for advanced high school students.

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