Ask a New Question
Menu

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, 6)
left parenthesis negative 3 comma 6 right parenthesis

(3, 9)

1 answer

To solve the system of equations using elimination, we start with the given equations:

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

We can multiply Equation 2 by 2 so that the coefficients of \( x \) will align for elimination:

\[ 2(-2x + 4y) = 2(30) \] This simplifies to:

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

Now, we will add Equation 1 and Equation 3 together:

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

This simplifies to:

\[ 0x + 2y = 12 \] We can now solve for \( y \):

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

Next, we substitute \( y = 6 \) back into one of the original equations to find \( x \). Let's use Equation 2:

\[ -2x + 4(6) = 30 \] This simplifies to:

\[ -2x + 24 = 30 \]

Subtracting 24 from both sides gives us:

\[ -2x = 6 \] So,

\[ x = -3 \]

Thus, the solution to the system of equations is:

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

This corresponds to the option:

\[ \text{(−3, 6)} \]