We need to solve 8 systems of linear equations with two variables, $x$ and $y$.

1. Problem Description

We need to solve 8 systems of linear equations with two variables, $x$ and $y$.

2. Solution Steps

1. System 1:

$-5x + y = -7$

$-3x - 2y = -12$

Multiply the first equation by 2:

$-10x + 2y = -14$

Add the modified first equation to the second equation:

$(-10x + 2y) + (-3x - 2y) = -14 + (-12)$

$-13x = -26$

$x = 2$

Substitute $x = 2$ into the first original equation:

$-5(2) + y = -7$

$-10 + y = -7$

$y = 3$

2. System 2:

$-2x + 6y = 6$

$-7x + 8y = -5$

Multiply the first equation by 7:

$-14x + 42y = 42$

Multiply the second equation by 2:

$-14x + 16y = -10$

Subtract the second modified equation from the first modified equation:

$(-14x + 42y) - (-14x + 16y) = 42 - (-10)$

$26y = 52$

$y = 2$

Substitute $y = 2$ into the first original equation:

$-2x + 6(2) = 6$

$-2x + 12 = 6$

$-2x = -6$

$x = 3$

3. System 3:

$-5x - y = 21$

$-4x + y = 6$

Add the two equations:

$(-5x - y) + (-4x + y) = 21 + 6$

$-9x = 27$

$x = -3$

Substitute $x = -3$ into the second equation:

$-4(-3) + y = 6$

$12 + y = 6$

$y = -6$

4. System 4:

$y = -3x$

$4x - 2y = -20$

Substitute $y = -3x$ into the second equation:

$4x - 2(-3x) = -20$

$4x + 6x = -20$

$10x = -20$

$x = -2$

Substitute $x = -2$ into the first equation:

$y = -3(-2)$

$y = 6$

5. System 5:

$x = 3y + 1$

$2x + 4y = 12$

Substitute $x = 3y + 1$ into the second equation:

$2(3y + 1) + 4y = 12$

$6y + 2 + 4y = 12$

$10y = 10$

$y = 1$

Substitute $y = 1$ into the first equation:

$x = 3(1) + 1$

$x = 4$

6. System 6:

$-5x - 8y = 17$

$2x - 7y = -17$

Multiply the first equation by 2:

$-10x - 16y = 34$

Multiply the second equation by 5:

$10x - 35y = -85$

Add the two modified equations:

$(-10x - 16y) + (10x - 35y) = 34 + (-85)$

$-51y = -51$

$y = 1$

Substitute $y = 1$ into the second equation:

$2x - 7(1) = -17$

$2x - 7 = -17$

$2x = -10$

$x = -5$

7. System 7:

$x + 9y = -1$

$2x + 4y = 5$

Multiply the first equation by -2:

$-2x - 18y = 2$

Add the modified first equation to the second equation:

$(-2x - 18y) + (2x + 4y) = 2 + 5$

$-14y = 7$

$y = -1/2 = -0.5$

Substitute $y = -1/2$ into the first equation:

$x + 9(-1/2) = -1$

$x - 9/2 = -1$

$x = -1 + 9/2 = -2/2 + 9/2 = 7/2 = 3.5$

8. System 8:

$x - y = 11$

$3x + 10y = -6$

Multiply the first equation by -3:

$-3x + 3y = -33$

Add the modified first equation to the second equation:

$(-3x + 3y) + (3x + 10y) = -33 + (-6)$

$13y = -39$

$y = -3$

Substitute $y = -3$ into the first equation:

$x - (-3) = 11$

$x + 3 = 11$

$x = 8$

3. Final Answer

1. $x = 2, y = 3$

2. $x = 3, y = 2$

3. $x = -3, y = -6$

4. $x = -2, y = 6$

5. $x = 4, y = 1$

6. $x = -5, y = 1$

7. $x = 7/2, y = -1/2$

8. $x = 8, y = -3$