Solve the system of two equations using elimination with multiplication.

Question

4x−6y=−48
−2x+4y=30 (1 point)
Responses

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

(3, 9)
left parenthesis 3 comma 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'll first write down the equations:

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

We can eliminate one of the variables by making the coefficients of \(x\) or \(y\) equal. Let's eliminate \(x\) by multiplying the second equation by 2:

1. \( 4x - 6y = -48 \) (keep this as is)
2. \( -2x + 4y = 30 \) (multiply by 2) \[ -4x + 8y = 60 \]

Now our system of equations looks like this:

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

Next, we add these two equations together:

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

Simplifying gives:

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

Now, we can solve for \(y\):

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

Now that we have \(y\), we substitute \(y = 6\) back into one of the original equations to solve for \(x\). We'll use the first equation:

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

Thus, the solution to the system of equations is \(x = -3\) and \(y = 6\). Therefore, the solution in coordinate form is:

\[ (-3, 6) \]

So the response is:

(−3, 6)