Question

Solve the system of two equations using elimination with multiplication.
4x−6y=−48

−2x+4y=30
(1 point)
Responses

(−21,−6)
left parenthesis negative 21 comma negative 6 right parenthesis

(−33,−9)
left parenthesis negative 33 comma negative 9 right parenthesis

(3, 9)
left parenthesis 3 comma 9 right parenthesis

(−3, 6)

Answer 1

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

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

We'll first manipulate the equations so that when we add or subtract them, one of the variables will cancel.

Let’s eliminate \( x \) by making the coefficients of \( x \) in both equations the same. We can multiply Equation 2 by 2, which will give both equations a coefficient of \( -4 \) for \( x \):

Multiply Equation 2 by 2:

\[
-2x + 4y = 30 \quad \Rightarrow \quad -4x + 8y = 60 \quad \text{(Equation 3)}
\]

Now, we rewrite our system of equations:

1. \( 4x - 6y = -48 \) (Equation 1)
2. \( -4x + 8y = 60 \) (Equation 3)

Next, we can add Equation 1 and Equation 3 together:

\[
(4x - 6y) + (-4x + 8y) = -48 + 60
\]
\[
4x - 4x - 6y + 8y = 12
\]
\[
2y = 12
\]
\[
y = 6
\]

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

\[
4x - 6(6) = -48
\]
\[
4x - 36 = -48
\]
\[
4x = -48 + 36
\]
\[
4x = -12
\]
\[
x = -3
\]

Thus, the solution for the system of equations is:

\[
\boxed{(-3, 6)}
\]

Verifying, we can substitute \( x = -3 \) and \( y = 6 \) back into both original equations to check:

For Equation 1:
\[
4(-3) - 6(6) = -12 - 36 = -48 \quad \text{(True)}
\]

For Equation 2:
\[
-2(-3) + 4(6) = 6 + 24 = 30 \quad \text{(True)}
\]

Both equations hold true, confirming that the solution is correct. The final answer is:

\[
\boxed{(-3, 6)}
\]