Math worksheet for solving systems of equations using the substitution method, featuring 12 problems with two equations each.

To solve the systems of equations using the substitution method, we will follow these steps for each problem:

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.

---

Problem 1:

\[
\begin{aligned}
1. & \quad x = 2y + 1 \\
& \quad 4x - 5y = 13
\end{aligned}
\]

#### Step 1: Substitute \( x = 2y + 1 \) into the second equation.
\[
4x - 5y = 13
\]
\[
4(2y + 1) - 5y = 13
\]

#### Step 2: Simplify and solve for \( y \).
\[
4(2y + 1) - 5y = 13
\]
\[
8y + 4 - 5y = 13
\]
\[
3y + 4 = 13
\]
\[
3y = 9
\]
\[
y = 3
\]

#### Step 3: Substitute \( y = 3 \) back into \( x = 2y + 1 \).
\[
x = 2(3) + 1
\]
\[
x = 6 + 1
\]
\[
x = 7
\]

#### Solution:
\[
(x, y) = (7, 3)
\]

---

Problem 2:

\[
\begin{aligned}
2. & \quad y = -3x - 6 \\
& \quad -5x - 8y = -28
\end{aligned}
\]

#### Step 1: Substitute \( y = -3x - 6 \) into the second equation.
\[
-5x - 8y = -28
\]
\[
-5x - 8(-3x - 6) = -28
\]

#### Step 2: Simplify and solve for \( x \).
\[
-5x - 8(-3x - 6) = -28
\]
\[
-5x + 24x + 48 = -28
\]
\[
19x + 48 = -28
\]
\[
19x = -76
\]
\[
x = -4
\]

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

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

---

Problem 3:

\[
\begin{aligned}
3. & \quad y = 2x - 20 \\
& \quad 8x + 7y = 80
\end{aligned}
\]

#### Step 1: Substitute \( y = 2x - 20 \) into the second equation.
\[
8x + 7y = 80
\]
\[
8x + 7(2x - 20) = 80
\]

#### Step 2: Simplify and solve for \( x \).
\[
8x + 7(2x - 20) = 80
\]
\[
8x + 14x - 140 = 80
\]
\[
22x - 140 = 80
\]
\[
22x = 220
\]
\[
x = 10
\]

#### Step 3: Substitute \( x = 10 \) back into \( y = 2x - 20 \).
\[
y = 2(10) - 20
\]
\[
y = 20 - 20
\]
\[
y = 0
\]

#### Solution:
\[
(x, y) = (10, 0)
\]

---

Problem 4:

\[
\begin{aligned}
4. & \quad x = -13y - 2 \\
& \quad x = 7y + 18
\end{aligned}
\]

#### Step 1: Set the two expressions for \( x \) equal to each other.
\[
-13y - 2 = 7y + 18
\]

#### Step 2: Solve for \( y \).
\[
-13y - 2 = 7y + 18
\]
\[
-13y - 7y = 18 + 2
\]
\[
-20y = 20
\]
\[
y = -1
\]

#### Step 3: Substitute \( y = -1 \) back into either equation for \( x \). Using \( x = -13y - 2 \):
\[
x = -13(-1) - 2
\]
\[
x = 13 - 2
\]
\[
x = 11
\]

#### Solution:
\[
(x, y) = (11, -1)
\]

---

Problem 5:

\[
\begin{aligned}
5. & \quad 2y = 4x \\
& \quad -3x + y = -4
\end{aligned}
\]

#### Step 1: Solve the first equation for \( y \).
\[
2y = 4x
\]
\[
y = 2x
\]

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

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

#### Solution:
\[
(x, y) = (4, 8)
\]

---

Problem 6:

\[
\begin{aligned}
6. & \quad 3x = -y \\
& \quad 10x + 4y = -6
\end{aligned}
\]

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

#### Step 2: Substitute \( y = -3x \) into the second equation.
\[
10x + 4y = -6
\]
\[
10x + 4(-3x) = -6
\]
\[
10x - 12x = -6
\]
\[
-2x = -6
\]
\[
x = 3
\]

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

#### Solution:
\[
(x, y) = (3, -9)
\]

---

Problem 7:

\[
\begin{aligned}
7. & \quad y = x + 16 \\
& \quad 10x + 8y = -88
\end{aligned}
\]

#### Step 1: Substitute \( y = x + 16 \) into the second equation.
\[
10x + 8y = -88
\]
\[
10x + 8(x + 16) = -88
\]

#### Step 2: Simplify and solve for \( x \).
\[
10x + 8(x + 16) = -88
\]
\[
10x + 8x + 128 = -88
\]
\[
18x + 128 = -88
\]
\[
18x = -216
\]
\[
x = -12
\]

#### Step 3: Substitute \( x = -12 \) back into \( y = x + 16 \).
\[
y = -12 + 16
\]
\[
y = 4
\]

#### Solution:
\[
(x, y) = (-12, 4)
\]

---

Problem 8:

\[
\begin{aligned}
8. & \quad x = 16 - 3y \\
& \quad 4x + 12y = 96
\end{aligned}
\]

#### Step 1: Substitute \( x = 16 - 3y \) into the second equation.
\[
4x + 12y = 96
\]
\[
4(16 - 3y) + 12y = 96
\]

#### Step 2: Simplify and solve for \( y \).
\[
4(16 - 3y) + 12y = 96
\]
\[
64 - 12y + 12y = 96
\]
\[
64 = 96
\]

This is a contradiction, so there is no solution.

#### Solution:
\[
\text{No solution}
\]

---

Problem 9:

\[
\begin{aligned}
9. & \quad x = \frac{1}{2}y - 7 \\
& \quad 3y = 42
\end{aligned}
\]

#### Step 1: Solve the second equation for \( y \).
\[
3y = 42
\]
\[
y = 14
\]

#### Step 2: Substitute \( y = 14 \) into \( x = \frac{1}{2}y - 7 \).
\[
x = \frac{1}{2}(14) - 7
\]
\[
x = 7 - 7
\]
\[
x = 0
\]

#### Solution:
\[
(x, y) = (0, 14)
\]

---

Problem 10:

\[
\begin{aligned}
10. & \quad x = -y \\
& \quad -3x - 3y = 0
\end{aligned}
\]

#### Step 1: Substitute \( x = -y \) into the second equation.
\[
-3x - 3y = 0
\]
\[
-3(-y) - 3y = 0
\]
\[
3y - 3y = 0
\]
\[
0 = 0
\]

This is an identity, so there are infinitely many solutions.

#### Solution:
\[
\text{Infinitely many solutions}
\]

---

Problem 11:

\[
\begin{aligned}
11. & \quad 4x = 8y - 52 \\
& \quad 5x - y = 34
\end{aligned}
\]

#### Step 1: Solve the first equation for \( x \).
\[
4x = 8y - 52
\]
\[
x = 2y - 13
\]

#### Step 2: Substitute \( x = 2y - 13 \) into the second equation.
\[
5x - y = 34
\]
\[
5(2y - 13) - y = 34
\]

#### Step 3: Simplify and solve for \( y \).
\[
5(2y - 13) - y = 34
\]
\[
10y - 65 - y = 34
\]
\[
9y - 65 = 34
\]
\[
9y = 99
\]
\[
y = 11
\]

#### Step 4: Substitute \( y = 11 \) back into \( x = 2y - 13 \).
\[
x = 2(11) - 13
\]
\[
x = 22 - 13
\]
\[
x = 9
\]

#### Solution:
\[
(x, y) = (9, 11)
\]

---

Problem 12:

\[
\begin{aligned}
12. & \quad 3y = 12 - 6x \\
& \quad -15x - 4y = -2
\end{aligned}
\]

#### Step 1: Solve the first equation for \( y \).
\[
3y = 12 - 6x
\]
\[
y = 4 - 2x
\]

#### Step 2: Substitute \( y = 4 - 2x \) into the second equation.
\[
-15x - 4y = -2
\]
\[
-15x - 4(4 - 2x) = -2
\]

#### Step 3: Simplify and solve for \( x \).
\[
-15x - 4(4 - 2x) = -2
\]
\[
-15x - 16 + 8x = -2
\]
\[
-7x - 16 = -2
\]
\[
-7x = 14
\]
\[
x = -2
\]

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

#### Solution:
\[
(x, y) = (-2, 8)
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad (7, 3) \\
2. & \quad (-4, 6) \\
3. & \quad (10, 0) \\
4. & \quad (11, -1) \\
5. & \quad (4, 8) \\
6. & \quad (3, -9) \\
7. & \quad (-12, 4) \\
8. & \quad \text{No solution} \\
9. & \quad (0, 14) \\
10. & \quad \text{Infinitely many solutions} \\
11. & \quad (9, 11) \\
12. & \quad (-2, 8)
\end{aligned}
}
\]