Question

1. $x + 9y = -8$

$-x - 8y = 7$

1. $7x - 7y = -14$

$-2x + 7y = -16$

1. $x + 7y = 3$

$4x + 7y = -9$

Explanation:

Let's solve problem 8: the system of equations \(

$$\begin{cases} x + 7y = 3 \\ 4x + 7y = -9 \end{cases}$$

\)

Step 1: Subtract the first equation from the second

Subtract \( x + 7y = 3 \) from \( 4x + 7y = -9 \) to eliminate \( y \).
\( (4x + 7y) - (x + 7y) = -9 - 3 \)
Simplify: \( 4x + 7y - x - 7y = -12 \)
Which gives: \( 3x = -12 \)

Step 2: Solve for \( x \)

Divide both sides of \( 3x = -12 \) by 3:
\( x = \frac{-12}{3} = -4 \)

Step 3: Substitute \( x = -4 \) into the first equation

Substitute \( x = -4 \) into \( x + 7y = 3 \):
\( -4 + 7y = 3 \)

Step 4: Solve for \( y \)

Add 4 to both sides: \( 7y = 3 + 4 = 7 \)
Divide both sides by 7: \( y = \frac{7}{7} = 1 \)

Answer:

The solution to the system is \( x = -4 \), \( y = 1 \)