If `x=2a^(2)+3b^(2)-5ab, y=b^(2)-3a^(2)+7ab and z=6a^(2)-b^(2)+ab`, find:
`x-y+z`

Answer

To solve the problem, we need to find the expression \( x - y + z \) using the given values of \( x \), \( y \), and \( z \).

### Step-by-Step Solution:

1. **Write down the expressions for \( x \), \( y \), and \( z \)**:
\[ x = 2a^2 + 3b^2 - 5ab \]
\[ y = b^2 - 3a^2 + 7ab \]
\[ z = 6a^2 - b^2 + ab \]

2. **Calculate \( x - y \)**:
\[ x - y = (2a^2 + 3b^2 - 5ab) - (b^2 - 3a^2 + 7ab) \]
Distributing the negative sign:
\[ = 2a^2 + 3b^2 - 5ab - b^2 + 3a^2 - 7ab \]
Now, combine like terms:
- For \( a^2 \): \( 2a^2 + 3a^2 = 5a^2 \)
- For \( b^2 \): \( 3b^2 - b^2 = 2b^2 \)
- For \( ab \): \( -5ab - 7ab = -12ab \)

Thus, we have:
\[ x - y = 5a^2 + 2b^2 - 12ab \]

3. **Now add \( z \) to \( x - y \)**:
\[ x - y + z = (5a^2 + 2b^2 - 12ab) + (6a^2 - b^2 + ab) \]
Combine the terms:
- For \( a^2 \): \( 5a^2 + 6a^2 = 11a^2 \)
- For \( b^2 \): \( 2b^2 - b^2 = b^2 \)
- For \( ab \): \( -12ab + ab = -11ab \)

Therefore, we have:
\[ x - y + z = 11a^2 + b^2 - 11ab \]

### Final Answer:
\[ x - y + z = 11a^2 + b^2 - 11ab \]