Question

1. $-x + 11y = -9$ $-10x - y = 21$ 10) $12x + 9y = 0$ $3x + 5y = 11$ $(-4)$ 11) $-4x + 10y = -4$ $(-10x + y = -10)$ $(-16)$ 12) $8x + 8y = 8$ $-7x - 16y = 2$ 13) $4x + 10y = -20$ $-3x - 5y = 5$ 14) $4x - 9y = 17$ $2x + 18y = -14$ 15) $14x + y = 9$ $7x - 3y = 22$ 16) $-4x - 9y = 5$ $-10x - 18y = -10$

Explanation:

Let's solve problem 10: \(

$$\begin{cases} 12x + 9y = 0 \\ 3x + 5y = 11 \end{cases}$$

\) (multiplying the second equation by -4 as per the note \( (3x + 5y = 11)(-4) \))

Step 1: Multiply the second equation by -4

To eliminate \( x \), we multiply the second equation \( 3x + 5y = 11 \) by -4.
\( -4(3x + 5y) = -4 \times 11 \)
\( -12x - 20y = -44 \)

Step 2: Add the first equation and the new second equation

The first equation is \( 12x + 9y = 0 \). Add it to \( -12x - 20y = -44 \):
\( (12x + 9y) + (-12x - 20y) = 0 + (-44) \)
\( 12x - 12x + 9y - 20y = -44 \)
\( -11y = -44 \)

Step 3: Solve for \( y \)

Divide both sides by -11:
\( y = \frac{-44}{-11} = 4 \)

Step 4: Substitute \( y = 4 \) into the first equation to find \( x \)

Substitute \( y = 4 \) into \( 12x + 9y = 0 \):
\( 12x + 9(4) = 0 \)
\( 12x + 36 = 0 \)
Subtract 36 from both sides:
\( 12x = -36 \)
Divide by 12:
\( x = \frac{-36}{12} = -3 \)

Answer:

\( x = -3 \), \( y = 4 \)