Question

Solve the system of two equations using elimination with multiplication. 4x-6y=-48 -2x+4y = 30 (1 point) ○ (3, 9) ○ (-33,-9) ○ (-21, -6) ○ (-3, 6)

Answer 1

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

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

To eliminate one variable, we will manipulate both equations. We can multiply the second equation by 2 to align the \(x\) coefficients:

Multiply the second equation by 2:
\[
2(-2x + 4y) = 2(30)
\]
This gives us:
\[
-4x + 8y = 60 \quad \text{(3)}
\]

Now we have the transformed system of equations:
1) \(4x - 6y = -48\)
3) \(-4x + 8y = 60\)

Next, we can add equations (1) and (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 \(y = 6\), we can substitute this value back into one of the original equations to find \(x\). Let's substitute into equation (1):

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

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

\[
(-3, 6)
\]

So the correct answer is \(( -3, 6 )\).