) 2x + 8y = 20					2)         x = 5 
              y = 2                                                   2x + y = 10 








3) 5x - 2y = 3 					4) 2y + x = -15
             y = 2x					            x = 3y







	 
5) 4x + 7y = 19					6)          y = 6x + 11
    	    y = x + 9					    2y - 4x = 14

Let's solve each system of equations individually.

---

### 1) System:  
\[
\begin{aligned}
2x + 8y &= 20 \\
x &= 5 \\
y &= 2 \\
\end{aligned}
\]
Substituting \( x = 5 \) and \( y = 2 \) into the first equation:
\[
2(5) + 8(2) = 20 \\
10 + 16 = 26 \quad (\text{False})
\]
So, this system has no solution as the first equation does not hold with the given values of \( x \) and \( y \).

---

### 2) System:  
\[
\begin{aligned}
x &= 5 \\
2x + y &= 10
\end{aligned}
\]
Substitute \( x = 5 \) into the second equation:
\[
2(5) + y = 10 \\
10 + y = 10 \\
y = 0
\]
So, the solution for this system is \( x = 5 \) and \( y = 0 \).

---

### 3) System:  
\[
\begin{aligned}
5x - 2y &= 3 \\
y &= 2x
\end{aligned}
\]
Substitute \( y = 2x \) into the first equation:
\[
5x - 2(2x) = 3 \\
5x - 4x = 3 \\
x = 3
\]
Substitute \( x = 3 \) back into \( y = 2x \):
\[
y = 2(3) = 6
\]
So, the solution for this system is \( x = 3 \) and \( y = 6 \).

---

### 4) System:  
\[
\begin{aligned}
2y + x &= -15 \\
x &= 3y
\end{aligned}
\]
Substitute \( x = 3y \) into the first equation:
\[
2y + 3y = -15 \\
5y = -15 \\
y = -3
\]
Substitute \( y = -3 \) back into \( x = 3y \):
\[
x = 3(-3) = -9
\]
So, the solution for this system is \( x = -9 \) and \( y = -3 \).

---

### 5) System:  
\[
\begin{aligned}
4x + 7y &= 19 \\
y &= x + 9
\end{aligned}
\]
Substitute \( y = x + 9 \) into the first equation:
\[
4x + 7(x + 9) = 19 \\
4x + 7x + 63 = 19 \\
11x + 63 = 19 \\
11x = 19 - 63 \\
11x = -44 \\
x = -4
\]
Substitute \( x = -4 \) back into \( y = x + 9 \):
\[
y = -4 + 9 = 5
\]
So, the solution for this system is \( x = -4 \) and \( y = 5 \).

---

### 6) System:  
\[
\begin{aligned}
y &= 6x + 11 \\
2y - 4x &= 14
\end{aligned}
\]
Substitute \( y = 6x + 11 \) into the second equation:
\[
2(6x + 11) - 4x = 14 \\
12x + 22 - 4x = 14 \\
8x + 22 = 14 \\
8x = 14 - 22 \\
8x = -8 \\
x = -1
\]
Substitute \( x = -1 \) back into \( y = 6x + 11 \):
\[
y = 6(-1) + 11 = -6 + 11 = 5
\]
So, the solution for this system is \( x = -1 \) and \( y = 5 \).

---

### Summary of Solutions:
1. No solution.
2. \( x = 5 \), \( y = 0 \)
3. \( x = 3 \), \( y = 6 \)
4. \( x = -9 \), \( y = -3 \)
5. \( x = -4 \), \( y = 5 \)
6. \( x = -1 \), \( y = 5 \)

---

Would you like more details on any of these solutions, or have any questions?

Here are 5 related questions to expand on this topic:
1. How do you solve systems of equations graphically?
2. What are the different methods for solving systems of equations?
3. Can a system of linear equations have infinite solutions?
4. How can substitution and elimination methods be compared?
5. What does it mean for a system of equations to be consistent or inconsistent?

**Tip**: Always check your solutions by substituting them back into the original equations to verify their correctness!

2x + 8y = 20, x = 5, y = 2; 2x + y = 10; 5x - 2y = 3, y = 2x; 2y + x = -15, x = 3y; 4x + 7y = 19, y = x + 9; y = 6x + 11, 2y - 4x = 14

Learn how to solve systems of linear equations using substitution and elimination methods. This detailed guide covers a variety of systems, including inconsistent and consistent solutions. Perfect for students in Grades 9-12 looking to master algebra.

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