Math worksheet for solving systems of equations using the substitution method.

To solve each system of equations by substitution, we will follow these steps:

1. Solve one of the equations for one variable in terms of the other.
2. Substitute this expression into the other equation to solve for the remaining variable.
3. Use the value found to solve for the other variable.

Let's solve each system step by step.

---

System ①:

\[
\begin{cases}
y = 6x - 11 \\
-2x - 3y = -7
\end{cases}
\]

#### Step 1: Substitute \( y = 6x - 11 \) into the second equation.
\[
-2x - 3(6x - 11) = -7
\]

#### Step 2: Simplify and solve for \( x \).
\[
-2x - 18x + 33 = -7
\]
\[
-20x + 33 = -7
\]
\[
-20x = -7 - 33
\]
\[
-20x = -40
\]
\[
x = 2
\]

#### Step 3: Substitute \( x = 2 \) back into \( y = 6x - 11 \).
\[
y = 6(2) - 11
\]
\[
y = 12 - 11
\]
\[
y = 1
\]

#### Solution:
\[
(x, y) = (2, 1)
\]

---

System ②:

\[
\begin{cases}
x + 3y = 6 \\
2x + 8y = -12
\end{cases}
\]

#### Step 1: Solve the first equation for \( x \).
\[
x = 6 - 3y
\]

#### Step 2: Substitute \( x = 6 - 3y \) into the second equation.
\[
2(6 - 3y) + 8y = -12
\]

#### Step 3: Simplify and solve for \( y \).
\[
12 - 6y + 8y = -12
\]
\[
12 + 2y = -12
\]
\[
2y = -12 - 12
\]
\[
2y = -24
\]
\[
y = -12
\]

#### Step 4: Substitute \( y = -12 \) back into \( x = 6 - 3y \).
\[
x = 6 - 3(-12)
\]
\[
x = 6 + 36
\]
\[
x = 42
\]

#### Solution:
\[
(x, y) = (42, -12)
\]

---

System ③:

\[
\begin{cases}
x + 3y = 12 \\
2x + y = 6
\end{cases}
\]

#### Step 1: Solve the second equation for \( y \).
\[
y = 6 - 2x
\]

#### Step 2: Substitute \( y = 6 - 2x \) into the first equation.
\[
x + 3(6 - 2x) = 12
\]

#### Step 3: Simplify and solve for \( x \).
\[
x + 18 - 6x = 12
\]
\[
-5x + 18 = 12
\]
\[
-5x = 12 - 18
\]
\[
-5x = -6
\]
\[
x = \frac{6}{5}
\]

#### Step 4: Substitute \( x = \frac{6}{5} \) back into \( y = 6 - 2x \).
\[
y = 6 - 2\left(\frac{6}{5}\right)
\]
\[
y = 6 - \frac{12}{5}
\]
\[
y = \frac{30}{5} - \frac{12}{5}
\]
\[
y = \frac{18}{5}
\]

#### Solution:
\[
(x, y) = \left(\frac{6}{5}, \frac{18}{5}\right)
\]

---

System ④:

\[
\begin{cases}
y = -2x - 9 \\
y = -5x - 21
\end{cases}
\]

#### Step 1: Set the two expressions for \( y \) equal to each other.
\[
-2x - 9 = -5x - 21
\]

#### Step 2: Solve for \( x \).
\[
-2x + 5x = -21 + 9
\]
\[
3x = -12
\]
\[
x = -4
\]

#### Step 3: Substitute \( x = -4 \) back into \( y = -2x - 9 \).
\[
y = -2(-4) - 9
\]
\[
y = 8 - 9
\]
\[
y = -1
\]

#### Solution:
\[
(x, y) = (-4, -1)
\]

---

System ⑤:

\[
\begin{cases}
-2x + 4y = -16 \\
y = x - 2
\end{cases}
\]

#### Step 1: Substitute \( y = x - 2 \) into the first equation.
\[
-2x + 4(x - 2) = -16
\]

#### Step 2: Simplify and solve for \( x \).
\[
-2x + 4x - 8 = -16
\]
\[
2x - 8 = -16
\]
\[
2x = -16 + 8
\]
\[
2x = -8
\]
\[
x = -4
\]

#### Step 3: Substitute \( x = -4 \) back into \( y = x - 2 \).
\[
y = -4 - 2
\]
\[
y = -6
\]

#### Solution:
\[
(x, y) = (-4, -6)
\]

---

System ⑥:

\[
\begin{cases}
2x - 3y = -1 \\
y = x - 1
\end{cases}
\]

#### Step 1: Substitute \( y = x - 1 \) into the first equation.
\[
2x - 3(x - 1) = -1
\]

#### Step 2: Simplify and solve for \( x \).
\[
2x - 3x + 3 = -1
\]
\[
-x + 3 = -1
\]
\[
-x = -1 - 3
\]
\[
-x = -4
\]
\[
x = 4
\]

#### Step 3: Substitute \( x = 4 \) back into \( y = x - 1 \).
\[
y = 4 - 1
\]
\[
y = 3
\]

#### Solution:
\[
(x, y) = (4, 3)
\]

---

System ⑦:

\[
\begin{cases}
y = 2x - 15 \\
y = 5x
\end{cases}
\]

#### Step 1: Set the two expressions for \( y \) equal to each other.
\[
2x - 15 = 5x
\]

#### Step 2: Solve for \( x \).
\[
2x - 5x = 15
\]
\[
-3x = 15
\]
\[
x = -5
\]

#### Step 3: Substitute \( x = -5 \) back into \( y = 5x \).
\[
y = 5(-5)
\]
\[
y = -25
\]

#### Solution:
\[
(x, y) = (-5, -25)
\]

---

System ⑧:

\[
\begin{cases}
2x + 4y = 4 \\
y = x - 2
\end{cases}
\]

#### Step 1: Substitute \( y = x - 2 \) into the first equation.
\[
2x + 4(x - 2) = 4
\]

#### Step 2: Simplify and solve for \( x \).
\[
2x + 4x - 8 = 4
\]
\[
6x - 8 = 4
\]
\[
6x = 4 + 8
\]
\[
6x = 12
\]
\[
x = 2
\]

#### Step 3: Substitute \( x = 2 \) back into \( y = x - 2 \).
\[
y = 2 - 2
\]
\[
y = 0
\]

#### Solution:
\[
(x, y) = (2, 0)
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{① } (2, 1) \\
&\text{② } (42, -12) \\
&\text{③ } \left(\frac{6}{5}, \frac{18}{5}\right) \\
&\text{④ } (-4, -1) \\
&\text{⑤ } (-4, -6) \\
&\text{⑥ } (4, 3) \\
&\text{⑦ } (-5, -25) \\
&\text{⑧ } (2, 0)
\end{aligned}
}
\]