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Solve the system of two equations using elimination with multiplication.

4x−6y=−48

−2x+4y=30
(1 point)
Responses

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

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

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

(−3, 6)

1 answer

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)

First, let's manipulate these equations to eliminate one of the variables. We can multiply Equation 2 by 2 to align the coefficients of \(x\):

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

This simplifies to:

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

Now we have a new system of equations:

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

Now, we can add Equation 1 and Equation 3 together to eliminate \(x\):

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

This simplifies to:

\[ (4x - 4x) + (-6y + 8y) = 12 \] \[ 0 + 2y = 12 \] \[ 2y = 12 \] \[ y = 6 \]

Now that we have \(y\), we can substitute it back into one of the original equations to find \(x\). Let's use Equation 2:

\[ -2x + 4(6) = 30 \] \[ -2x + 24 = 30 \] \[ -2x = 30 - 24 \] \[ -2x = 6 \] \[ x = -3 \]

So, the solution to the system of equations is:

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

Thus, the final answer is:

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