Solve the following systems of equations using the substitution method: [specific systems of equations were provided in the solution]

Here are the solutions to the systems of equations using the substitution method.

1.
Given the system:

$$x + \frac{1}{2}y = 5 \quad (1)$$ $$y = 4x \quad (2)$$

Step 1: Substitute equation (2) into equation (1).

$$x + \frac{1}{2}(4x) = 5$$ $$x + 2x = 5$$ $$3x = 5$$ $$x = \frac{5}{3}$$

Step 2: Substitute $x=\frac{5}{3}$ into equation (2).

$$y = 4\left(\frac{5}{3}\right)$$ $$y = \frac{20}{3}$$

The solution is $(x, y) = \left(\frac{5}{3}, \frac{20}{3}\right)$.

$$\boxed{x=\frac{5}{3}, y=\frac{20}{3}}$$

2.
Given the system:

$$x = \frac{1}{2}y + 2 \quad (1)$$ $$x + y = 10 \quad (2)$$

Step 1: Substitute equation (1) into equation (2).

$$\left(\frac{1}{2}y + 2\right) + y = 10$$ $$\frac{3}{2}y + 2 = 10$$ $$\frac{3}{2}y = 8$$ $$y = 8 \times \frac{2}{3}$$ $$y = \frac{16}{3}$$

Step 2: Substitute $y=\frac{16}{3}$ into equation (1).

$$x = \frac{1}{2}\left(\frac{16}{3}\right) + 2$$ $$x = \frac{8}{3} + \frac{6}{3}$$ $$x = \frac{14}{3}$$

The solution is $(x, y) = \left(\frac{14}{3}, \frac{16}{3}\right)$.

$$\boxed{x=\frac{14}{3}, y=\frac{16}{3}}$$

3.
Given the system:

$$y = \frac{1}{2}x - 1 \quad (1)$$ $$x + y = 7 \quad (2)$$

Step 1: Substitute equation (1) into equation (2).

$$x + \left(\frac{1}{2}x - 1\right) = 7$$ $$\frac{3}{2}x - 1 = 7$$ $$\frac{3}{2}x = 8$$ $$x = 8 \times \frac{2}{3}$$ $$x = \frac{16}{3}$$

Step 2: Substitute $x=\frac{16}{3}$ into equation (1).

$$y = \frac{1}{2}\left(\frac{16}{3}\right) - 1$$ $$y = \frac{8}{3} - \frac{3}{3}$$ $$y = \frac{5}{3}$$

The solution is $(x, y) = \left(\frac{16}{3}, \frac{5}{3}\right)$.

$$\boxed{x=\frac{16}{3}, y=\frac{5}{3}}$$

4.
Given the system:

$$x = \frac{2}{5}y \quad (1)$$ $$x + y = 9 \quad (2)$$

Step 1: Substitute equation (1) into equation (2).

$$\frac{2}{5}y + y = 9$$ $$\frac{7}{5}y = 9$$ $$y = 9 \times \frac{5}{7}$$ $$y = \frac{45}{7}$$

Step 2: Substitute $y=\frac{45}{7}$ into equation (1).

$$x = \frac{2}{5}\left(\frac{45}{7}\right)$$ $$x = \frac{18}{7}$$

The solution is $(x, y) = \left(\frac{18}{7}, \frac{45}{7}\right)$.

$$\boxed{x=\frac{18}{7}, y=\frac{45}{7}}$$

5.
Given the system:

$$y = \frac{3}{4}x \quad (1)$$ $$x + y = 14 \quad (2)$$

Step 1: Substitute equation (1) into equation (2).

$$x + \frac{3}{4}x = 14$$ $$\frac{7}{4}x = 14$$ $$x = 14 \times \frac{4}{7}$$ $$x = 8$$

Step 2: Substitute $x=8$ into equation (1).

$$y = \frac{3}{4}(8)$$ $$y = 6$$

The solution is $(x, y) = (8, 6)$.

$$\boxed{x=8, y=6}$$

6.
Given the system:

$$x = \frac{1}{2}y + 3 \quad (1)$$ $$x - y = 1 \quad (2)$$

Step 1: Substitute equation (1) into equation (2).

$$\left(\frac{1}{2}y + 3\right) - y = 1$$ $$-\frac{1}{2}y + 3 = 1$$ $$-\frac{1}{2}y = -2$$ $$y = 4$$

Step 2: Substitute $y=4$ into equation (1).

$$x = \frac{1}{2}(4) + 3$$ $$x = 2 + 3$$ $$x = 5$$

The solution is $(x, y) = (5, 4)$.

$$\boxed{x=5, y=4}$$

7.
Given the system:

$$y = \frac{2}{3}x + 1 \quad (1)$$ $$x + y = 11 \quad (2)$$

Step 1: Substitute equation (1) into equation (2).

$$x + \left(\frac{2}{3}x + 1\right) = 11$$ $$\frac{5}{3}x + 1 = 11$$ $$\frac{5}{3}x = 10$$ $$$$