https://file.aiquickdraw.com/mathbot/file/172779293271919rj8zj8.jpg

The image presents two systems of equations that need to be solved:

### First system:
1. \( 2x - 4y = 20 \)
2. \( x = 4 \)

### Second system:
1. \( y = 6x + 11 \)
2. \( 2x - 3y = 7 \)

Let's solve each system one by one:

### First system solution:
Given \( x = 4 \), substitute this into the first equation:
\[
2(4) - 4y = 20
\]
\[
8 - 4y = 20
\]
Now, solve for \( y \):
\[
-4y = 20 - 8
\]
\[
-4y = 12
\]
\[
y = -3
\]
Thus, the solution to the first system is \( x = 4 \) and \( y = -3 \).

### Second system solution:
From the first equation, \( y = 6x + 11 \). Substitute this into the second equation:
\[
2x - 3(6x + 11) = 7
\]
Distribute the \( -3 \):
\[
2x - 18x - 33 = 7
\]
Now, combine like terms:
\[
-16x - 33 = 7
\]
Solve for \( x \):
\[
-16x = 7 + 33
\]
\[
-16x = 40
\]
\[
x = -\frac{40}{16} = -\frac{5}{2}
\]
Now substitute \( x = -\frac{5}{2} \) into \( y = 6x + 11 \):
\[
y = 6\left(-\frac{5}{2}\right) + 11
\]
\[
y = -15 + 11 = -4
\]
Thus, the solution to the second system is \( x = -\frac{5}{2} \) and \( y = -4 \).

### Final answers:
1. First system: \( x = 4 \), \( y = -3 \)
2. Second system: \( x = -\frac{5}{2} \), \( y = -4 \)

Would you like further details on any step? Here are some related questions you might explore:

1. How can I verify the solution to a system of equations?
2. What are the methods for solving a system of linear equations?
3. How can substitution and elimination methods be compared in solving systems?
4. Can systems of equations have more than one solution?
5. What happens if the equations in a system are dependent or inconsistent?

**Tip:** Always substitute your solutions back into the original equations to ensure accuracy.

Solve the systems of equations: (1) 2x - 4y = 20, x = 4; (2) y = 6x + 11, 2x - 3y = 7.

Learn how to solve systems of linear equations using the substitution method. The problem involves two systems of equations with detailed steps, covering algebraic concepts and substitution techniques. Suitable for Grades 8-10 students.

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