Question

Solve the system of two equations using elimination with multiplication. 4x−6y=−48 −2x+4y=30 (1 point) Responses (3, 9) left parenthesis 3 comma 9 right parenthesis (−21,−6) left parenthesis negative 21 comma negative 6 right parenthesis (−3, 6) left parenthesis negative 3 comma 6 right parenthesis (−33,−9)

Answer 1

To solve the system of equations using elimination, we have the following two equations:

1. \( 4x - 6y = -48 \) (Equation 1)
2. \( -2x + 4y = 30 \) (Equation 2)

We can manipulate one or both of the equations so that when we add or subtract them, one variable will eliminate.

To do this, we can multiply the second equation (Equation 2) by 2, so that the coefficient of \(x\) in both equations will be the same (but will have opposite signs):

\[
-2x + 4y = 30 \quad \text{(Multiply by 2)}
\]
\[
-4x + 8y = 60 \quad \text{(New Equation 2)}
\]

Now we have the following equations:
1. \( 4x - 6y = -48 \)
2. \( -4x + 8y = 60 \)

Next, we add both equations together:

\[
(4x - 6y) + (-4x + 8y) = -48 + 60
\]
This simplifies to:
\[
(4x - 4x) + (-6y + 8y) = 12
\]
\[
0 + 2y = 12
\]

Now we can solve for \(y\):
\[
2y = 12 \quad \Rightarrow \quad y = 6
\]

Now that we have \(y\), we can substitute it back into either of the original equations to find \(x\). Let's use Equation 1:

\[
4x - 6(6) = -48
\]
\[
4x - 36 = -48
\]
Adding 36 to both sides:
\[
4x = -48 + 36
\]
\[
4x = -12
\]

Now, divide by 4:
\[
x = -3
\]

We find \(x = -3\) and \(y = 6\). Therefore, the solution to the system of equations is:

\[
(-3, 6)
\]

Thus, the correct response is \((-3, 6)\).