Elementary algebra worksheet on solving systems of linear equations using substitution, with step-by-step examples and exercises.

Problem: Solving a System of Two Linear Equations in Two Variables by Substitution

The task is to solve each system of linear equations using the substitution method. Let's go through each problem step by step.

---

#### Problem 1:
\[
\begin{aligned}
1) & \quad y = -4x + 16 \quad \text{(1)} \\
& \quad -3x + 8y = 23 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Substitute \( y = -4x + 16 \) from equation (1) into equation (2):
\[
-3x + 8(-4x + 16) = 23
\]
2. Simplify:
\[
-3x - 32x + 128 = 23
\]
\[
-35x + 128 = 23
\]
3. Solve for \( x \):
\[
-35x = 23 - 128
\]
\[
-35x = -105
\]
\[
x = \frac{-105}{-35} = 3
\]
4. Substitute \( x = 3 \) back into equation (1) to find \( y \):
\[
y = -4(3) + 16
\]
\[
y = -12 + 16 = 4
\]

Answer:
\[
\boxed{x = 3, y = 4}
\]

---

#### Problem 2:
\[
\begin{aligned}
2) & \quad -3x + 6y = -24 \quad \text{(1)} \\
& \quad y = 7x + 22 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Substitute \( y = 7x + 22 \) from equation (2) into equation (1):
\[
-3x + 6(7x + 22) = -24
\]
2. Simplify:
\[
-3x + 42x + 132 = -24
\]
\[
39x + 132 = -24
\]
3. Solve for \( x \):
\[
39x = -24 - 132
\]
\[
39x = -156
\]
\[
x = \frac{-156}{39} = -4
\]
4. Substitute \( x = -4 \) back into equation (2) to find \( y \):
\[
y = 7(-4) + 22
\]
\[
y = -28 + 22 = -6
\]

Answer:
\[
\boxed{x = -4, y = -6}
\]

---

#### Problem 3:
\[
\begin{aligned}
3) & \quad y = 5x + 5 \quad \text{(1)} \\
& \quad y = x + 5 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Since both equations are solved for \( y \), set them equal to each other:
\[
5x + 5 = x + 5
\]
2. Solve for \( x \):
\[
5x - x = 5 - 5
\]
\[
4x = 0
\]
\[
x = 0
\]
3. Substitute \( x = 0 \) back into either equation (1) or (2) to find \( y \):
\[
y = 5(0) + 5 = 5
\]

Answer:
\[
\boxed{x = 0, y = 5}
\]

---

#### Problem 4:
\[
\begin{aligned}
4) & \quad y = 4x + 22 \quad \text{(1)} \\
& \quad y = -4x - 18 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Set the two expressions for \( y \) equal to each other:
\[
4x + 22 = -4x - 18
\]
2. Solve for \( x \):
\[
4x + 4x = -18 - 22
\]
\[
8x = -40
\]
\[
x = \frac{-40}{8} = -5
\]
3. Substitute \( x = -5 \) back into either equation (1) or (2) to find \( y \):
\[
y = 4(-5) + 22 = -20 + 22 = 2
\]

Answer:
\[
\boxed{x = -5, y = 2}
\]

---

#### Problem 5:
\[
\begin{aligned}
5) & \quad x - 3y = -12 \quad \text{(1)} \\
& \quad 4x + 6y = -12 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Solve equation (1) for \( x \):
\[
x = 3y - 12
\]
2. Substitute \( x = 3y - 12 \) into equation (2):
\[
4(3y - 12) + 6y = -12
\]
3. Simplify:
\[
12y - 48 + 6y = -12
\]
\[
18y - 48 = -12
\]
4. Solve for \( y \):
\[
18y = -12 + 48
\]
\[
18y = 36
\]
\[
y = \frac{36}{18} = 2
\]
5. Substitute \( y = 2 \) back into \( x = 3y - 12 \) to find \( x \):
\[
x = 3(2) - 12 = 6 - 12 = -6
\]

Answer:
\[
\boxed{x = -6, y = 2}
\]

---

#### Problem 6:
\[
\begin{aligned}
6) & \quad 2x - 5y = 22 \quad \text{(1)} \\
& \quad x + 5y = -4 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Solve equation (2) for \( x \):
\[
x = -5y - 4
\]
2. Substitute \( x = -5y - 4 \) into equation (1):
\[
2(-5y - 4) - 5y = 22
\]
3. Simplify:
\[
-10y - 8 - 5y = 22
\]
\[
-15y - 8 = 22
\]
4. Solve for \( y \):
\[
-15y = 22 + 8
\]
\[
-15y = 30
\]
\[
y = \frac{30}{-15} = -2
\]
5. Substitute \( y = -2 \) back into \( x = -5y - 4 \) to find \( x \):
\[
x = -5(-2) - 4 = 10 - 4 = 6
\]

Answer:
\[
\boxed{x = 6, y = -2}
\]

---

#### Problem 7:
\[
\begin{aligned}
7) & \quad y = -4 \quad \text{(1)} \\
& \quad -3x - 6y = 15 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Substitute \( y = -4 \) from equation (1) into equation (2):
\[
-3x - 6(-4) = 15
\]
2. Simplify:
\[
-3x + 24 = 15
\]
3. Solve for \( x \):
\[
-3x = 15 - 24
\]
\[
-3x = -9
\]
\[
x = \frac{-9}{-3} = 3
\]

Answer:
\[
\boxed{x = 3, y = -4}
\]

---

#### Problem 8:
\[
\begin{aligned}
8) & \quad 2x + 4y = -10 \quad \text{(1)} \\
& \quad 7x + 8y = -23 \quad \text{(2)}
\end{aligned}
\]

Solution:
1. Solve equation (1) for \( y \):
\[
2x + 4y = -10
\]
\[
4y = -10 - 2x
\]
\[
y = \frac{-10 - 2x}{4} = -\frac{5}{2} - \frac{x}{2}
\]
2. Substitute \( y = -\frac{5}{2} - \frac{x}{2} \) into equation (2):
\[
7x + 8\left(-\frac{5}{2} - \frac{x}{2}\right) = -23
\]
3. Simplify:
\[
7x + 8\left(-\frac{5}{2}\right) + 8\left(-\frac{x}{2}\right) = -23
\]
\[
7x - 20 - 4x = -23
\]
\[
3x - 20 = -23
\]
4. Solve for \( x \):
\[
3x = -23 + 20
\]
\[
3x = -3
\]
\[
x = \frac{-3}{3} = -1
\]
5. Substitute \( x = -1 \) back into \( y = -\frac{5}{2} - \frac{x}{2} \) to find \( y \):
\[
y = -\frac{5}{2} - \frac{-1}{2} = -\frac{5}{2} + \frac{1}{2} = -\frac{4}{2} = -2
\]

Answer:
\[
\boxed{x = -1, y = -2}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \quad x = 3, y = 4 \\
2) & \quad x = -4, y = -6 \\
3) & \quad x = 0, y = 5 \\
4) & \quad x = -5, y = 2 \\
5) & \quad x = -6, y = 2 \\
6) & \quad x = 6, y = -2 \\
7) & \quad x = 3, y = -4 \\
8) & \quad x = -1, y = -2
\end{aligned}
}
\]