Worksheet displaying six systems of linear equations to solve.

To solve the given system of equations, we will address each pair of equations step by step. Let's go through them one by one.

---

Problem 1:

\[
\begin{aligned}
1. & \quad y = 5x - 11 \\
& \quad y = -2x + 10
\end{aligned}
\]

#### Solution:
Since both equations are already solved for \( y \), we can set them equal to each other:
\[
5x - 11 = -2x + 10
\]

Solve for \( x \):
\[
5x + 2x = 10 + 11
\]
\[
7x = 21
\]
\[
x = 3
\]

Now substitute \( x = 3 \) back into one of the original equations to find \( y \). Using \( y = 5x - 11 \):
\[
y = 5(3) - 11
\]
\[
y = 15 - 11
\]
\[
y = 4
\]

Thus, the solution is:
\[
\boxed{(3, 4)}
\]

---

Problem 2:

\[
\begin{aligned}
2. & \quad 7x - 3y = -24 \\
& \quad 3x - 7y = -16
\end{aligned}
\]

#### Solution:
We will use the method of elimination. First, we need to eliminate one of the variables. Let's eliminate \( y \). To do this, we make the coefficients of \( y \) in both equations equal in magnitude but opposite in sign.

Multiply the first equation by 7 and the second equation by 3:
\[
7(7x - 3y) = 7(-24) \quad \Rightarrow \quad 49x - 21y = -168
\]
\[
3(3x - 7y) = 3(-16) \quad \Rightarrow \quad 9x - 21y = -48
\]

Now subtract the second equation from the first:
\[
(49x - 21y) - (9x - 21y) = -168 - (-48)
\]
\[
49x - 9x - 21y + 21y = -168 + 48
\]
\[
40x = -120
\]
\[
x = -3
\]

Now substitute \( x = -3 \) back into one of the original equations to find \( y \). Using \( 7x - 3y = -24 \):
\[
7(-3) - 3y = -24
\]
\[
-21 - 3y = -24
\]
\[
-3y = -24 + 21
\]
\[
-3y = -3
\]
\[
y = 1
\]

Thus, the solution is:
\[
\boxed{(-3, 1)}
\]

---

Problem 3:

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

#### Solution:
Substitute \( y = -3x + 6 \) into the second equation \( 2x - 3y = 4 \):
\[
2x - 3(-3x + 6) = 4
\]
\[
2x + 9x - 18 = 4
\]
\[
11x - 18 = 4
\]
\[
11x = 22
\]
\[
x = 2
\]

Now substitute \( x = 2 \) back into \( y = -3x + 6 \):
\[
y = -3(2) + 6
\]
\[
y = -6 + 6
\]
\[
y = 0
\]

Thus, the solution is:
\[
\boxed{(2, 0)}
\]

---

Problem 4:

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

#### Solution:
Substitute \( y = x - 4 \) into the second equation \( x + 2y = 1 \):
\[
x + 2(x - 4) = 1
\]
\[
x + 2x - 8 = 1
\]
\[
3x - 8 = 1
\]
\[
3x = 9
\]
\[
x = 3
\]

Now substitute \( x = 3 \) back into \( y = x - 4 \):
\[
y = 3 - 4
\]
\[
y = -1
\]

Thus, the solution is:
\[
\boxed{(3, -1)}
\]

---

Problem 5:

\[
\begin{aligned}
5. & \quad 4x - 5y = -10 \\
& \quad 4x - 5y = -5
\end{aligned}
\]

#### Solution:
Notice that both equations are identical except for the constant term. This means the two lines are parallel and never intersect. Therefore, there is no solution.

Thus, the solution is:
\[
\boxed{\text{No solution}}
\]

---

Problem 6:

\[
\begin{aligned}
6. & \quad 3y = 4x - 3 \\
& \quad 8x - 6y = 6
\end{aligned}
\]

#### Solution:
First, rewrite the first equation in standard form:
\[
3y = 4x - 3 \quad \Rightarrow \quad 4x - 3y = 3
\]

Now we have:
\[
\begin{aligned}
& \quad 4x - 3y = 3 \\
& \quad 8x - 6y = 6
\end{aligned}
\]

Notice that the second equation is a multiple of the first equation. Specifically, the second equation is exactly twice the first equation. This means the two equations represent the same line, and there are infinitely many solutions.

Thus, the solution is:
\[
\boxed{\text{Infinitely many solutions}}
\]

---

Final Answers:

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