Question

Solve the system of equations below using elimination.

2x+3y=12


3x−6y=−24


(1 point)
Responses

(-12, -10)
(-12, -10)

(4, 0)
(4, 0)

(0, -8)
(0, -8)

(0, 4)

Answer 1

To solve the system of equations using the elimination method, we start with the two equations provided:

1. \(2x + 3y = 12\) (Equation 1)
2. \(3x - 6y = -24\) (Equation 2)

First, we will manipulate the equations to eliminate one variable. Let's eliminate \(y\). To do this, we can multiply Equation 1 by 2 in order to make the coefficients of \(y\) in both equations match.

Multiplying Equation 1 by 2:

\[
2(2x + 3y) = 2(12) \\
4x + 6y = 24 \quad (Equation 3)
\]

Now we have:

1. \(4x + 6y = 24\) (Equation 3)
2. \(3x - 6y = -24\) (Equation 2)

Next, we can add Equation 3 and Equation 2 together to eliminate \(y\):

\[
(4x + 6y) + (3x - 6y) = 24 + (-24) \\
4x + 3x + 6y - 6y = 0 \\
7x = 0
\]

Now, solve for \(x\):

\[
x = 0
\]

Now that we have found \(x\), we can substitute \(x = 0\) back into one of the original equations to solve for \(y\). We can use Equation 1:

\[
2(0) + 3y = 12 \\
3y = 12
\]

Now solve for \(y\):

\[
y = \frac{12}{3} = 4
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (0, 4)
\]

The correct response is:

\((0, 4)\)