Systems Of Equations Substitution Maze Worksheet - Free Printable

To solve the "Systems of Equations Substitution Maze," we need to solve each system of equations using the substitution method. The solutions will guide us through the maze from "Start" to "Finish." Let's go step by step.

Step 1: Solve the first system at "Start"

The system is:
\[
x = -2y
\]
\[
4x + 3y = 10
\]

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

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

So, the solution is \( (4, -2) \).

Step 2: Follow the path to the next system

From \( (4, -2) \), the next system is:
\[
x = 3y + 8
\]
\[
x + 4y = 15
\]

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

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

So, the solution is \( (11, 1) \).

Step 3: Continue solving systems along the path

We will continue this process for each system until we reach the "Finish" box. Here are the solutions for the remaining systems:

1. Next system:
\[
y = 12x - 5
\]
\[
y = 3x + 4
\]

#### Set the equations equal to each other:
\[
12x - 5 = 3x + 4
\]
\[
12x - 3x = 4 + 5
\]
\[
9x = 9
\]
\[
x = 1
\]

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

So, the solution is \( (1, 7) \).

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

So, the solution is \( (-6, 1) \).

3. Next 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 \( y = 4x + 15 \):
\[
y = 4(-2) + 15
\]
\[
y = -8 + 15
\]
\[
y = 7
\]

So, the solution is \( (-2, 7) \).

4. Next 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 \( y = -5x - 9 \):
\[
y = -5(-3) - 9
\]
\[
y = 15 - 9
\]
\[
y = 6
\]

So, the solution is \( (-3, 6) \).

5. Next 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 \( x = 5y + 4 \):
\[
x = 5(-2) + 4
\]
\[
x = -10 + 4
\]
\[
x = -6
\]

So, the solution is \( (-6, 2) \).

6. Next 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 \( x = -y - 5 \):
\[
x = -(-1) - 5
\]
\[
x = 1 - 5
\]
\[
x = -4
\]

So, the solution is \( (-4, -1) \).

Final Answer:

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

\[
\boxed{(-4, -1)}
\]