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Slove this problem in step by step equation 

The problem in the image shows a system of equations to solve. I will break down each equation step by step:

### Equations from the image:

1. \( C = 6Z - 4Y + P \)
2. \( C = 9Z - 7Y \)
3. \( P = 2Y - 3Z \)

### Step-by-Step Solution:

#### Step 1: Set the first two equations equal to each other
Since both equations represent \( C \), we can set them equal:

\[
6Z - 4Y + P = 9Z - 7Y
\]

#### Step 2: Substitute the value of \( P \)
From the third equation, we know \( P = 2Y - 3Z \). Substitute this into the equation from Step 1:

\[
6Z - 4Y + (2Y - 3Z) = 9Z - 7Y
\]

#### Step 3: Simplify the equation
Now, simplify the terms:

\[
6Z - 4Y + 2Y - 3Z = 9Z - 7Y
\]

This simplifies to:

\[
(6Z - 3Z) + (-4Y + 2Y) = 9Z - 7Y
\]
\[
3Z - 2Y = 9Z - 7Y
\]

#### Step 4: Combine like terms
Move all \( Z \)-terms to one side and \( Y \)-terms to the other:

\[
3Z - 9Z = -7Y + 2Y
\]
\[
-6Z = -5Y
\]

Simplify:

\[
Z = \frac{5}{6}Y
\]

#### Step 5: Substitute \( Z \) back into the third equation
Substitute \( Z = \frac{5}{6}Y \) into the equation for \( P \):

\[
P = 2Y - 3Z
\]
\[
P = 2Y - 3\left(\frac{5}{6}Y\right)
\]
\[
P = 2Y - \frac{15}{6}Y
\]
\[
P = 2Y - 2.5Y
\]
\[
P = -0.5Y
\]

#### Final Result:
- \( Z = \frac{5}{6}Y \)
- \( P = -0.5Y \)

Would you like further details on these steps or how to use these results? Here are 5 questions to explore further:

1. How do you apply substitution in a system of equations?
2. Can this system of equations be solved graphically?
3. What are the implications of solving \( Z \) in terms of \( Y \)?
4. How does the relationship between \( Z \) and \( Y \) affect \( P \)?
5. What other methods can be used to solve this system of equations?

**Tip:** When solving systems of equations, substituting one variable into another equation often simplifies the problem significantly!

Solve this problem step by step: C = 6Z - 4Y + P, C = 9Z - 7Y, P = 2Y - 3Z

Learn how to solve the system of equations involving variables C, Z, Y, and P step by step. This solution covers the substitution method and algebraic simplifications, ideal for high school algebra students (Grades 9-12).

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