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Let's solve the system of equations using substitution. The system is:

\[
y = -2x - 5 \tag{1}
\]
\[
-3x^2 + 4x - y = 8 \tag{2}
\]

### Step 1: Substitute \( y \) from equation (1) into equation (2).

Substitute \( y = -2x - 5 \) into equation (2):

\[
-3x^2 + 4x - (-2x - 5) = 8
\]

Simplify:

\[
-3x^2 + 4x + 2x + 5 = 8
\]
\[
-3x^2 + 6x + 5 = 8
\]

### Step 2: Solve for \( x \).

Subtract 8 from both sides:

\[
-3x^2 + 6x + 5 - 8 = 0
\]
\[
-3x^2 + 6x - 3 = 0
\]

Divide the entire equation by -3 to simplify:

\[
x^2 - 2x + 1 = 0
\]

This simplifies to:

\[
(x - 1)^2 = 0
\]

So, \( x = 1 \).

### Step 3: Substitute \( x = 1 \) into equation (1) to find \( y \).

Using \( y = -2x - 5 \):

\[
y = -2(1) - 5 = -2 - 5 = -7
\]

Thus, the solution is \( (x, y) = (1, -7) \).

### Final Answer:
The solution is \( \boxed{(1, -7)} \).

Would you like further details on any part of the solution?

Here are 5 related questions you might find helpful:
1. How do you solve quadratic equations using factoring?
2. What are other methods to solve a system of equations, like elimination?
3. How do you check if a solution is correct for a system of equations?
4. What are some real-world applications of solving systems of equations?
5. How do you solve a quadratic equation when it doesn't factor easily?

**Tip:** Always double-check your solution by plugging the values back into the original equations!

Solve the system by substitution: y = -2x - 5 and -3x^2 + 4x - y = 8.

Learn how to solve a system of equations involving a linear equation y = -2x - 5 and a quadratic equation -3x^2 + 4x - y = 8 using the substitution method. This step-by-step guide is suitable for students in Grades 9-11.

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