Linear Systems (A) Answers worksheet featuring eight systems of equations and their corresponding solutions.

The task involves solving systems of linear equations. Each system consists of two equations with two variables, and the goal is to find the values of the variables that satisfy both equations simultaneously. Below, I will explain the solution process for each system step by step.

---

System 1:

\[
\begin{aligned}
1. & \quad 3u + z = 15 \\
& \quad u + 2z = 10
\end{aligned}
\]

#### Step 1: Solve for one variable in terms of the other.
From the second equation:
\[
u + 2z = 10 \implies u = 10 - 2z
\]

#### Step 2: Substitute \( u = 10 - 2z \) into the first equation.
\[
3u + z = 15 \implies 3(10 - 2z) + z = 15
\]
Simplify:
\[
30 - 6z + z = 15 \implies 30 - 5z = 15
\]
Solve for \( z \):
\[
-5z = 15 - 30 \implies -5z = -15 \implies z = 3
\]

#### Step 3: Substitute \( z = 3 \) back into \( u = 10 - 2z \).
\[
u = 10 - 2(3) = 10 - 6 = 4
\]

#### Solution:
\[
u = 4, z = 3
\]

---

System 2:

\[
\begin{aligned}
2. & \quad u + 6y = 32 \\
& \quad u + 3y = 17
\end{aligned}
\]

#### Step 1: Subtract the second equation from the first to eliminate \( u \).
\[
(u + 6y) - (u + 3y) = 32 - 17
\]
Simplify:
\[
6y - 3y = 15 \implies 3y = 15 \implies y = 5
\]

#### Step 2: Substitute \( y = 5 \) into the second equation.
\[
u + 3y = 17 \implies u + 3(5) = 17
\]
Simplify:
\[
u + 15 = 17 \implies u = 2
\]

#### Solution:
\[
u = 2, y = 5
\]

---

System 3:

\[
\begin{aligned}
3. & \quad 3c + 4u = 33 \\
& \quad 6c + 3u = 36
\end{aligned}
\]

#### Step 1: Simplify the second equation by dividing by 3.
\[
6c + 3u = 36 \implies 2c + u = 12
\]

#### Step 2: Solve for \( u \) in terms of \( c \) from the simplified second equation.
\[
2c + u = 12 \implies u = 12 - 2c
\]

#### Step 3: Substitute \( u = 12 - 2c \) into the first equation.
\[
3c + 4u = 33 \implies 3c + 4(12 - 2c) = 33
\]
Simplify:
\[
3c + 48 - 8c = 33 \implies -5c + 48 = 33
\]
Solve for \( c \):
\[
-5c = 33 - 48 \implies -5c = -15 \implies c = 3
\]

#### Step 4: Substitute \( c = 3 \) back into \( u = 12 - 2c \).
\[
u = 12 - 2(3) = 12 - 6 = 6
\]

#### Solution:
\[
c = 3, u = 6
\]

---

System 4:

\[
\begin{aligned}
4. & \quad 6u + v = 18 \\
& \quad 5u + 2v = 22
\end{aligned}
\]

#### Step 1: Solve for \( v \) in terms of \( u \) from the first equation.
\[
6u + v = 18 \implies v = 18 - 6u
\]

#### Step 2: Substitute \( v = 18 - 6u \) into the second equation.
\[
5u + 2v = 22 \implies 5u + 2(18 - 6u) = 22
\]
Simplify:
\[
5u + 36 - 12u = 22 \implies -7u + 36 = 22
\]
Solve for \( u \):
\[
-7u = 22 - 36 \implies -7u = -14 \implies u = 2
\]

#### Step 3: Substitute \( u = 2 \) back into \( v = 18 - 6u \).
\[
v = 18 - 6(2) = 18 - 12 = 6
\]

#### Solution:
\[
u = 2, v = 6
\]

---

System 5:

\[
\begin{aligned}
5. & \quad 2a + 2x = 18 \\
& \quad a + 3x = 17
\end{aligned}
\]

#### Step 1: Simplify the first equation by dividing by 2.
\[
2a + 2x = 18 \implies a + x = 9
\]

#### Step 2: Solve for \( a \) in terms of \( x \) from the simplified first equation.
\[
a + x = 9 \implies a = 9 - x
\]

#### Step 3: Substitute \( a = 9 - x \) into the second equation.
\[
a + 3x = 17 \implies (9 - x) + 3x = 17
\]
Simplify:
\[
9 - x + 3x = 17 \implies 9 + 2x = 17
\]
Solve for \( x \):
\[
2x = 17 - 9 \implies 2x = 8 \implies x = 4
\]

#### Step 4: Substitute \( x = 4 \) back into \( a = 9 - x \).
\[
a = 9 - 4 = 5
\]

#### Solution:
\[
a = 5, x = 4
\]

---

System 6:

\[
\begin{aligned}
6. & \quad 5a + 2v = 32 \\
& \quad 6a + 6v = 42
\end{aligned}
\]

#### Step 1: Simplify the second equation by dividing by 6.
\[
6a + 6v = 42 \implies a + v = 7
\]

#### Step 2: Solve for \( v \) in terms of \( a \) from the simplified second equation.
\[
a + v = 7 \implies v = 7 - a
\]

#### Step 3: Substitute \( v = 7 - a \) into the first equation.
\[
5a + 2v = 32 \implies 5a + 2(7 - a) = 32
\]
Simplify:
\[
5a + 14 - 2a = 32 \implies 3a + 14 = 32
\]
Solve for \( a \):
\[
3a = 32 - 14 \implies 3a = 18 \implies a = 6
\]

#### Step 4: Substitute \( a = 6 \) back into \( v = 7 - a \).
\[
v = 7 - 6 = 1
\]

#### Solution:
\[
a = 6, v = 1
\]

---

System 7:

\[
\begin{aligned}
7. & \quad 2b + v = 13 \\
& \quad b + v = 8
\end{aligned}
\]

#### Step 1: Subtract the second equation from the first to eliminate \( v \).
\[
(2b + v) - (b + v) = 13 - 8
\]
Simplify:
\[
2b - b = 5 \implies b = 5
\]

#### Step 2: Substitute \( b = 5 \) into the second equation.
\[
b + v = 8 \implies 5 + v = 8
\]
Solve for \( v \):
\[
v = 8 - 5 \implies v = 3
\]

#### Solution:
\[
b = 5, v = 3
\]

---

System 8:

\[
\begin{aligned}
8. & \quad 3a + 5u = 17 \\
& \quad 2a + u = 9
\end{aligned}
\]

#### Step 1: Solve for \( u \) in terms of \( a \) from the second equation.
\[
2a + u = 9 \implies u = 9 - 2a
\]

#### Step 2: Substitute \( u = 9 - 2a \) into the first equation.
\[
3a + 5u = 17 \implies 3a + 5(9 - 2a) = 17
\]
Simplify:
\[
3a + 45 - 10a = 17 \implies -7a + 45 = 17
\]
Solve for \( a \):
\[
-7a = 17 - 45 \implies -7a = -28 \implies a = 4
\]

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

#### Solution:
\[
a = 4, u = 1
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad u = 4, z = 3 \\
2. & \quad u = 2, y = 5 \\
3. & \quad c = 3, u = 6 \\
4. & \quad u = 2, v = 6 \\
5. & \quad a = 5, x = 4 \\
6. & \quad a = 6, v = 1 \\
7. & \quad b = 5, v = 3 \\
8. & \quad a = 4, u = 1
\end{aligned}
}
\]