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

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

(−3, 6)
left parenthesis negative 3 comma 6 right parenthesis

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

(3, 9)

Answer 1

To solve the system of equations using elimination, we'll manipulate the equations to eliminate one of the variables. The given equations are:

1) \(4x - 6y = -48\)

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

First, we can multiply the second equation by 2 to make the coefficients of \(x\) in both equations match:

Multiplying the second equation by 2 gives:

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

Now we rewrite the system of equations:

1) \(4x - 6y = -48\)

2) \(-4x + 8y = 60\)

Next, we add the two equations together to eliminate \(x\):

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

This simplifies to:

\[
0x + 2y = 12
\]

This means:

\[
2y = 12
\]

Dividing both sides by 2 gives us:

\[
y = 6
\]

Now that we have \(y\), we can substitute this value back into one of the original equations to find \(x\). We'll use the first equation:

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

This simplifies to:

\[
4x - 36 = -48
\]

Adding 36 to both sides:

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

Dividing both sides by 4 gives:

\[
x = -3
\]

Thus, the solution to the system of equations is:

\((-3, 6)\)

So the correct response is:

**(−3, 6)**