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Let's solve the Systems of Equations Substitution Maze step by step.

We are given a maze where each box contains a system of two equations. We must solve each system using substitution, find the solution (x, y), and use that to determine which path to follow from Start to Finish.

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🟩 Step 1: Start at the "Start" Box

Start Box:

$$
\begin{align*}
x &= -2y \quad \text{(1)}\\
4x + 3y &= 10 \quad \text{(2)}
\end{align*}
$$

Substitute equation (1) into equation (2):

$$
4(-2y) + 3y = 10 \\
-8y + 3y = 10 \\
-5y = 10 \\
y = -2
$$

Now plug back into (1):

$$
x = -2(-2) = 4
$$

So the solution is: (4, -2)

✔ This matches the arrow pointing to the next box: (4, -2)

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🟨 Next Box: (4, -2)

This leads to the box:

$$
\begin{align*}
x &= 3y + 8 \quad \text{(1)}\\
x + 4y &= 15 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
(3y + 8) + 4y = 15 \\
7y + 8 = 15 \\
7y = 7 \\
y = 1
$$

Now plug into (1):

$$
x = 3(1) + 8 = 11
$$

So the solution is: (11, 1)

✔ Arrow points to (11, 1)

---

🟨 Next Box: (11, 1)

Next system:

$$
\begin{align*}
y &= 12x - 5 \quad \text{(1)}\\
y &= 3x + 4 \quad \text{(2)}
\end{align*}
$$

Set them equal:

$$
12x - 5 = 3x + 4 \\
9x = 9 \\
x = 1
$$

Then $ y = 3(1) + 4 = 7 $

Solution: (1, 7)

✔ Arrow points to (1, 7)

---

🟨 Next Box: (1, 7)

System:

$$
\begin{align*}
x &= -5y - 1 \quad \text{(1)}\\
y &= 2x + 13 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
y = 2(-5y - 1) + 13 \\
y = -10y - 2 + 13 \\
y = -10y + 11 \\
11y = 11 \\
y = 1
$$

Now plug into (1):

$$
x = -5(1) - 1 = -6
$$

Solution: (-6, 1)

✔ Arrow points to (-6, 1)

---

🟨 Next Box: (-6, 1)

System:

$$
\begin{align*}
y &= 4x + 15 \quad \text{(1)}\\
-5x - y &= 3 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
-5x - (4x + 15) = 3 \\
-5x - 4x - 15 = 3 \\
-9x = 18 \\
x = -2
$$

Then $ y = 4(-2) + 15 = -8 + 15 = 7 $

Solution: (-2, 7)

✔ Arrow points to (-2, 7)

---

🟨 Next Box: (-2, 7)

System:

$$
\begin{align*}
y &= -4x - 5 \quad \text{(1)}\\
3x + y &= -3 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
3x + (-4x - 5) = -3 \\
- x - 5 = -3 \\
- x = 2 \\
x = -2
$$

Then $ y = -4(-2) - 5 = 8 - 5 = 3 $

Solution: (-2, 3)

✔ Arrow points to (-2, 3)

---

🟨 Next Box: (-2, 3)

System:

$$
\begin{align*}
x &= -y + 5 \quad \text{(1)}\\
3x + 5y &= 15 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
3(-y + 5) + 5y = 15 \\
-3y + 15 + 5y = 15 \\
2y + 15 = 15 \\
2y = 0 \\
y = 0
$$

Then $ x = -0 + 5 = 5 $

Solution: (5, 0)

✔ Arrow points to (5, 0)

---

🟨 Next Box: (5, 0)

System:

$$
\begin{align*}
y &= 6x + 14 \quad \text{(1)}\\
2x + y &= 6 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
2x + (6x + 14) = 6 \\
8x + 14 = 6 \\
8x = -8 \\
x = -1
$$

Then $ y = 6(-1) + 14 = -6 + 14 = 8 $

Solution: (-1, 8)

✔ Arrow points to (-1, 8)

---

🟨 Next Box: (-1, 8)

System:

$$
\begin{align*}
-2x + y &= 12 \quad \text{(1)}\\
y &= -5x - 9 \quad \text{(2)}
\end{align*}
$$

Substitute (2) into (1):

$$
-2x + (-5x - 9) = 12 \\
-7x - 9 = 12 \\
-7x = 21 \\
x = -3
$$

Then $ y = -5(-3) - 9 = 15 - 9 = 6 $

Solution: (-3, 6)

✔ Arrow points to (-3, 6)

---

🟨 Next Box: (-3, 6)

System:

$$
\begin{align*}
y &= 3x \quad \text{(1)}\\
y &= 7x + 16 \quad \text{(2)}
\end{align*}
$$

Set equal:

$$
3x = 7x + 16 \\
-4x = 16 \\
x = -4
$$

Then $ y = 3(-4) = -12 $

Solution: (-4, -12)

✔ Arrow points to (-4, -12)

---

🟨 Next Box: (-4, -12)

System:

$$
\begin{align*}
9x - 5y &= -9 \quad \text{(1)}\\
y &= 4x - 7 \quad \text{(2)}
\end{align*}
$$

Substitute (2) into (1):

$$
9x - 5(4x - 7) = -9 \\
9x - 20x + 35 = -9 \\
-11x + 35 = -9 \\
-11x = -44 \\
x = 4
$$

Then $ y = 4(4) - 7 = 16 - 7 = 9 $

Solution: (4, 9)

✔ Arrow points to (4, 9)

---

🟨 Next Box: (4, 9)

System:

$$
\begin{align*}
x &= -4y - 9 \quad \text{(1)}\\
3x + 6y &= -3 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
3(-4y - 9) + 6y = -3 \\
-12y - 27 + 6y = -3 \\
-6y - 27 = -3 \\
-6y = 24 \\
y = -4
$$

Then $ x = -4(-4) - 9 = 16 - 9 = 7 $

Solution: (7, -4)

✔ Arrow points to (7, -4)

---

🟨 Next Box: (7, -4)

System:

$$
\begin{align*}
y &= 3x - 10 \quad \text{(1)}\\
3x + 4y &= 5 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
3x + 4(3x - 10) = 5 \\
3x + 12x - 40 = 5 \\
15x = 45 \\
x = 3
$$

Then $ y = 3(3) - 10 = 9 - 10 = -1 $

Solution: (3, -1)

✔ Arrow points to (3, -1)

---

🟨 Next Box: (3, -1)

System:

$$
\begin{align*}
x &= 5y + 4 \quad \text{(1)}\\
3x - 2y &= -14 \quad \text{(2)}
\end{align*}
$$

Substitute (1) into (2):

$$
3(5y + 4) - 2y = -14 \\
15y + 12 - 2y = -14 \\
13y + 12 = -14 \\
13y = -26 \\
y = -2
$$

Then $ x = 5(-2) + 4 = -10 + 4 = -6 $

Solution: (-6, 2)

✔ Arrow points to (-6, 2)

---

🟨 Next Box: (-6, 2)

System:

$$
\begin{align*}
x &= -y - 5 \quad \text{(1)}\\
x &= 2y - 2 \quad \text{(2)}
\end{align*}
$$

Set equal:

$$
-y - 5 = 2y - 2 \\
-5 + 2 = 2y + y \\
-3 = 3y \\
y = -1
$$

Then $ x = -(-1) - 5 = 1 - 5 = -4 $

Solution: (-4, -1)

✔ Arrow points to (-4, -1)

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🟩 Final Box: (-4, -1)

This leads directly to Finish ✔

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✔ Final Answer: Path from Start to Finish

Here is the correct path through the maze:

1. Start
2. → (4, -2)
3. → (11, 1)
4. → (1, 7)
5. → (-6, 1)
6. → (-2, 7)
7. → (-2, 3)
8. → (5, 0)
9. → (-1, 8)
10. → (-3, 6)
11. → (-4, -12)
12. → (4, 9)
13. → (7, -4)
14. → (3, -1)
15. → (-6, 2)
16. → (-4, -1)
17. → Finish

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🧠 Summary

You solved 14 systems of equations using substitution, each time finding the correct solution and following the path based on the answer.

The key steps were:
- Substitute one equation into the other.
- Solve for one variable.
- Plug back in to find the second.
- Match the solution to the labeled point and follow the path.

✔ All steps check out — you successfully navigated the maze!

Let me know if you'd like a visual diagram or printable version!