Solve systems of equations using substitution to navigate the maze from start to finish.

To solve the "Systems of Equations Substitution Maze," we need to solve each system of equations using the substitution method. The solution to each system will give us a coordinate pair (x, y). We then use these coordinates to navigate through the maze from "Start" to "Finish."

Step-by-Step Solution:

#### 1. Start:
- System:
\[
x = -2y
\]
\[
4x + 3y = 10
\]

- Substitute \( x = -2y \) into the second equation:
\[
4(-2y) + 3y = 10
\]
\[
-8y + 3y = 10
\]
\[
-5y = 10
\]
\[
y = -2
\]

- Substitute \( y = -2 \) back into \( x = -2y \):
\[
x = -2(-2) = 4
\]

- Solution: \((4, -2)\)

#### 2. Next Box:
- System:
\[
x = 3y + 8
\]
\[
x + 4y = 15
\]

- Substitute \( x = 3y + 8 \) into the second equation:
\[
(3y + 8) + 4y = 15
\]
\[
7y + 8 = 15
\]
\[
7y = 7
\]
\[
y = 1
\]

- Substitute \( y = 1 \) back into \( x = 3y + 8 \):
\[
x = 3(1) + 8 = 11
\]

- Solution: \((11, 1)\)

#### 3. Next Box:
- System:
\[
y = 12x - 5
\]
\[
y = 3x + 4
\]

- Since both equations are solved for \( y \), set them equal to each other:
\[
12x - 5 = 3x + 4
\]
\[
12x - 3x = 4 + 5
\]
\[
9x = 9
\]
\[
x = 1
\]

- Substitute \( x = 1 \) into \( y = 3x + 4 \):
\[
y = 3(1) + 4 = 7
\]

- Solution: \((1, 7)\)

#### 4. Next Box:
- System:
\[
x = -5y - 1
\]
\[
y = 2x + 13
\]

- Substitute \( y = 2x + 13 \) into \( x = -5y - 1 \):
\[
x = -5(2x + 13) - 1
\]
\[
x = -10x - 65 - 1
\]
\[
x = -10x - 66
\]
\[
x + 10x = -66
\]
\[
11x = -66
\]
\[
x = -6
\]

- Substitute \( x = -6 \) into \( y = 2x + 13 \):
\[
y = 2(-6) + 13 = -12 + 13 = 1
\]

- Solution: \((-6, 1)\)

#### 5. Next Box:
- System:
\[
y = 4x + 15
\]
\[
-5x - y = 3
\]

- Substitute \( y = 4x + 15 \) into the second equation:
\[
-5x - (4x + 15) = 3
\]
\[
-5x - 4x - 15 = 3
\]
\[
-9x - 15 = 3
\]
\[
-9x = 18
\]
\[
x = -2
\]

- Substitute \( x = -2 \) back into \( y = 4x + 15 \):
\[
y = 4(-2) + 15 = -8 + 15 = 7
\]

- Solution: \((-2, 7)\)

#### 6. Next Box:
- System:
\[
y = -4x - 5
\]
\[
3x + y = -3
\]

- Substitute \( y = -4x - 5 \) into the second equation:
\[
3x + (-4x - 5) = -3
\]
\[
3x - 4x - 5 = -3
\]
\[
-x - 5 = -3
\]
\[
-x = 2
\]
\[
x = -2
\]

- Substitute \( x = -2 \) back into \( y = -4x - 5 \):
\[
y = -4(-2) - 5 = 8 - 5 = 3
\]

- Solution: \((-2, 3)\)

#### 7. Next Box:
- System:
\[
x = -y + 5
\]
\[
3x + 5y = 15
\]

- Substitute \( x = -y + 5 \) into the second equation:
\[
3(-y + 5) + 5y = 15
\]
\[
-3y + 15 + 5y = 15
\]
\[
2y + 15 = 15
\]
\[
2y = 0
\]
\[
y = 0
\]

- Substitute \( y = 0 \) back into \( x = -y + 5 \):
\[
x = -0 + 5 = 5
\]

- Solution: \((5, 0)\)

#### 8. Next Box:
- System:
\[
y = 6x + 14
\]
\[
2x + y = 6
\]

- Substitute \( y = 6x + 14 \) into the second equation:
\[
2x + (6x + 14) = 6
\]
\[
2x + 6x + 14 = 6
\]
\[
8x + 14 = 6
\]
\[
8x = -8
\]
\[
x = -1
\]

- Substitute \( x = -1 \) back into \( y = 6x + 14 \):
\[
y = 6(-1) + 14 = -6 + 14 = 8
\]

- Solution: \((-1, 8)\)

#### 9. Next Box:
- System:
\[
-2x + y = 12
\]
\[
y = -5x - 9
\]

- Substitute \( y = -5x - 9 \) into the first equation:
\[
-2x + (-5x - 9) = 12
\]
\[
-2x - 5x - 9 = 12
\]
\[
-7x - 9 = 12
\]
\[
-7x = 21
\]
\[
x = -3
\]

- Substitute \( x = -3 \) back into \( y = -5x - 9 \):
\[
y = -5(-3) - 9 = 15 - 9 = 6
\]

- Solution: \((-3, 6)\)

#### 10. Next Box:
- System:
\[
y = 3x
\]
\[
y = 7x + 16
\]

- Set the equations equal to each other:
\[
3x = 7x + 16
\]
\[
3x - 7x = 16
\]
\[
-4x = 16
\]
\[
x = -4
\]

- Substitute \( x = -4 \) into \( y = 3x \):
\[
y = 3(-4) = -12
\]

- Solution: \((-4, -12)\)

#### 11. Next Box:
- System:
\[
9x - 5y = -9
\]
\[
y = 4x - 7
\]

- Substitute \( y = 4x - 7 \) into the first equation:
\[
9x - 5(4x - 7) = -9
\]
\[
9x - 20x + 35 = -9
\]
\[
-11x + 35 = -9
\]
\[
-11x = -44
\]
\[
x = 4
\]

- Substitute \( x = 4 \) back into \( y = 4x - 7 \):
\[
y = 4(4) - 7 = 16 - 7 = 9
\]

- Solution: \((4, 9)\)

#### 12. Next Box:
- System:
\[
x = -4y - 9
\]
\[
3x + 6y = -3
\]

- Substitute \( x = -4y - 9 \) into the second equation:
\[
3(-4y - 9) + 6y = -3
\]
\[
-12y - 27 + 6y = -3
\]
\[
-6y - 27 = -3
\]
\[
-6y = 24
\]
\[
y = -4
\]

- Substitute \( y = -4 \) back into \( x = -4y - 9 \):
\[
x = -4(-4) - 9 = 16 - 9 = 7
\]

- Solution: \((7, -4)\)

#### 13. Next Box:
- System:
\[
y = 3x - 10
\]
\[
3x + 4y = 5
\]

- Substitute \( y = 3x - 10 \) into the second equation:
\[
3x + 4(3x - 10) = 5
\]
\[
3x + 12x - 40 = 5
\]
\[
15x - 40 = 5
\]
\[
15x = 45
\]
\[
x = 3
\]

- Substitute \( x = 3 \) back into \( y = 3x - 10 \):
\[
y = 3(3) - 10 = 9 - 10 = -1
\]

- Solution: \((3, -1)\)

#### 14. Next Box:
- System:
\[
x = 5y + 4
\]
\[
3x - 2y = -14
\]

- Substitute \( x = 5y + 4 \) into the second equation:
\[
3(5y + 4) - 2y = -14
\]
\[
15y + 12 - 2y = -14
\]
\[
13y + 12 = -14
\]
\[
13y = -26
\]
\[
y = -2
\]

- Substitute \( y = -2 \) back into \( x = 5y + 4 \):
\[
x = 5(-2) + 4 = -10 + 4 = -6
\]

- Solution: \((-6, 2)\)

#### 15. Next Box:
- System:
\[
x = -y - 5
\]
\[
x = 2y - 2
\]

- Set the equations equal to each other:
\[
-y - 5 = 2y - 2
\]
\[
-y - 2y = -2 + 5
\]
\[
-3y = 3
\]
\[
y = -1
\]

- Substitute \( y = -1 \) into \( x = -y - 5 \):
\[
x = -(-1) - 5 = 1 - 5 = -4
\]

- Solution: \((-4, -1)\)

Final Answer:

The path from "Start" to "Finish" is:
\[
\boxed{(4, -2) \rightarrow (11, 1) \rightarrow (1, 7) \rightarrow (-6, 1) \rightarrow (-2, 7) \rightarrow (-2, 3) \rightarrow (5, 0) \rightarrow (4, 9) \rightarrow (7, -4) \rightarrow (3, -1) \rightarrow (-6, 2) \rightarrow (-4, -1)}
\]