To solve each system of equations, we will follow standard algebraic methods such as the substitution method or the elimination method. Here is the detailed step-by-step solution for each pair of equations:
### System 1:
[tex]\[
\begin{cases}
x + 3y = -26 \quad \text{(1)} \\
-x + 6y = -46 \quad \text{(2)}
\end{cases}
\][/tex]
1. Add equation (1) and (2) to eliminate [tex]\(x\)[/tex]:
[tex]\[
(x + 3y) + (-x + 6y) = -26 + (-46)
\][/tex]
[tex]\[
9y = -72
\][/tex]
[tex]\[
y = -8
\][/tex]
2. Substitute [tex]\(y = -8\)[/tex] back into equation (1):
[tex]\[
x + 3(-8) = -26
\][/tex]
[tex]\[
x - 24 = -26
\][/tex]
[tex]\[
x = -2
\][/tex]
The solution for System 1 is [tex]\((-2, -8)\)[/tex].
### System 2:
[tex]\[
\begin{cases}
5x + 9y = 29 \quad \text{(3)} \\
15x + 27y = -3 \quad \text{(4)}
\end{cases}
\][/tex]
1. Recognize that equation (4) is a multiple of equation (3):
[tex]\[
15x + 27y = 3(5x + 9y) = 3 \times 29 = 87
\][/tex]
But given as -3, which creates contradiction. This implies:
[tex]\[
15x + 27y = -3 \Rightarrow \text{Contradiction with } 3(5x + 9y) = 87
\][/tex]
There are no solutions for System 2, so the answer is [tex]\(\text{DNE}\)[/tex].
### System 3:
[tex]\[
\begin{cases}
3x + 9y = 66 \quad \text{(5)} \\
-3x + 2y = 0 \quad \text{(6)}
\end{cases}
\][/tex]
1. Add equation (5) and (6):
[tex]\[
3x + 9y + (-3x + 2y) = 66 + 0
\][/tex]
[tex]\[
11y = 66
\][/tex]
[tex]\[
y = 6
\][/tex]
2. Substitute [tex]\(y = 6\)[/tex] back into equation (6):
[tex]\[
-3x + 2(6) = 0
\][/tex]
[tex]\[
-3x + 12 = 0
\][/tex]
[tex]\[
-3x = -12
\][/tex]
[tex]\[
x = 4
\][/tex]
The solution for System 3 is [tex]\((4, 6)\)[/tex].
### System 4:
[tex]\[
\begin{cases}
7x - 9y = 9 \quad \text{(7)} \\
-14x + 18y = -18 \quad \text{(8)}
\end{cases}
\][/tex]
1. Notice that equation (8) is a multiple of equation (7):
[tex]\[
-14x + 18y = -2(7x - 9y) = -2 \times 9 = -18
\][/tex]
Since these equations are multiples of each other, they represent the same line. Therefore, there are infinitely many solutions that can be expressed as:
[tex]\[
7x - 9y = 9 \implies x = \frac{9y + 9}{7}
\][/tex]
The solution can be written as:
[tex]\[
(x, y) = \left( \frac{9y + 9}{7}, y \right)
\][/tex]
### Summary:
System 1: [tex]\((-2, -8)\)[/tex]
System 2: [tex]\(\text{DNE}\)[/tex]
System 3: [tex]\((4, 6)\)[/tex]
System 4: [tex]\(\left( \frac{9y + 9}{7}, y \right)\)[/tex] where [tex]\( y \)[/tex] is any real number.
