Question

solve the following system of equations. -9x - 8y = -20 3x + 2y = 2 x = \square y = \square

Explanation:

Step1: Multiply the second equation by 4

To eliminate \( y \), we multiply the second equation \( 3x + 2y = 2 \) by 4. This gives us \( 12x + 8y = 8 \).

Step2: Add the two equations

Now we add the first equation \( -9x - 8y = -20 \) and the new second equation \( 12x + 8y = 8 \). When we add them, the \( y \) terms cancel out:
\[

$$\begin{align*} (-9x - 8y) + (12x + 8y) &= -20 + 8 \\ -9x + 12x - 8y + 8y &= -12 \\ 3x &= -12 \end{align*}$$

\]

Step3: Solve for \( x \)

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

Step4: Substitute \( x = -4 \) into the second equation

Substitute \( x = -4 \) into \( 3x + 2y = 2 \):
\[
3(-4) + 2y = 2
\]

Step5: Solve for \( y \)

Simplify the left side: \( -12 + 2y = 2 \). Add 12 to both sides: \( 2y = 2 + 12 = 14 \). Then divide by 2: \( y = \frac{14}{2} = 7 \).

Answer:

\( x = -4 \)
\( y = 7 \)