Question

Solve the system by elimination.
-3x + 8y = -6
3x - 2y = -12

Answer 1

To solve the system of equations by elimination, we start with the two equations:

1. \(-3x + 8y = -6\) (Equation 1)
2. \(3x - 2y = -12\) (Equation 2)

To eliminate \(x\), we can add both equations together. We can do this by manipulating the coefficients to make them equal and opposite. Since the coefficient of \(x\) in Equation 1 is \(-3\) and in Equation 2 is \(3\), they are already opposites. We just add the equations as they are.

Let's add Equation 1 and Equation 2:

\[
-3x + 8y + 3x - 2y = -6 - 12
\]

Combining like terms on the left side gives:

\[
( -3x + 3x ) + ( 8y - 2y ) = -18
\]
\[
0 + 6y = -18
\]

Now, we solve for \(y\):

\[
6y = -18
\]
\[
y = \frac{-18}{6} = -3
\]

Now that we have \(y = -3\), we substitute this value back into one of the original equations to find \(x\). We'll use Equation 2:

\[
3x - 2(-3) = -12
\]
\[
3x + 6 = -12
\]

Next, we isolate \(x\):

\[
3x = -12 - 6
\]
\[
3x = -18
\]
\[
x = \frac{-18}{3} = -6
\]

We have the solution:

\[
x = -6, \quad y = -3
\]

Thus, the solution to the system of equations is \((x, y) = (-6, -3)\).