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The system of equations provided is:

\[
2x^2 - 3x - y = -5 \quad \text{(Equation 1)}
\]
\[
-x + y = 5 \quad \text{(Equation 2)}
\]

### Step 1: Solve the second equation for \( y \)
From the second equation, we can express \( y \) in terms of \( x \):
\[
y = x + 5
\]

### Step 2: Substitute \( y = x + 5 \) into the first equation
Substitute this into the first equation \( 2x^2 - 3x - y = -5 \):

\[
2x^2 - 3x - (x + 5) = -5
\]
Simplify this equation:

\[
2x^2 - 3x - x - 5 = -5
\]
\[
2x^2 - 4x = 0
\]

### Step 3: Solve for \( x \)
Factor the quadratic equation:

\[
2x(x - 2) = 0
\]

This gives two solutions for \( x \):

\[
x = 0 \quad \text{or} \quad x = 2
\]

### Step 4: Find \( y \) for each \( x \)
- If \( x = 0 \), substitute into \( y = x + 5 \):
  \[
  y = 0 + 5 = 5
  \]
  So, one solution is \( (0, 5) \).

- If \( x = 2 \), substitute into \( y = x + 5 \):
  \[
  y = 2 + 5 = 7
  \]
  So, another solution is \( (2, 7) \).

### Final Solutions:
The solutions are \( (0, 5) \) and \( (2, 7) \).

The correct answer from the given choices is:
- \( (0, 5) \)
- \( (2, 7) \)

Would you like more details or have any questions?

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Here are 5 related questions to expand on this:
1. How do we factor quadratic equations more generally?
2. What happens if the system of equations results in no solutions?
3. How can substitution be used in systems involving more variables?
4. Can elimination be used on non-linear systems, and how?
5. What methods exist for solving systems that are more complex than quadratic?

**Tip**: Always check both solutions in the original system to ensure they satisfy both equations.

Solve the system by elimination. Select the solution(s). 2x^2 - 3x - y = -5, -x + y = 5.

Learn how to solve a system of equations by elimination and substitution. This example features a quadratic equation 2x^2 - 3x - y = -5 and a linear equation -x + y = 5. Suitable for students in Grades 8-10.

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