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Question

Solving with Elimination Quick Check
3 of 53 of 5 Items

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

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

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

(−3, 6)

Answer 1

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

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

First, we can manipulate the equations to align them for elimination. To do this, let's multiply the second equation by 2 to make the coefficients of \(x\) match (but opposite in sign):

\[ 2(-2x + 4y) = 2(30) \]

This simplifies to:

\[ -4x + 8y = 60 \]

Now our system of equations looks like this:

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

Now we can add both equations together:

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

This simplifies to:

\[ 0x + 2y = 12 \]

Now, solving for \(y\):

\[ 2y = 12 \ y = 6 \]

Next, we substitute \(y = 6\) back into one of the original equations to find \(x\). We'll use the first equation:

\[ 4x - 6(6) = -48 \]

This simplifies to:

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

Thus, the solution to the system is:

\[ (x, y) = (-3, 6) \]

So, the correct answer from the provided options is:

\((-3, 6)\)