Question

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

Answer 1

To solve the system of equations:

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

we can use the elimination method. We need to make the coefficients of \(x\) or \(y\) equal in both equations.

Let's manipulate Equation 2 to help eliminate \(x\). First, we can multiply Equation 2 by 2 to make the coefficient of \(x\) equivalent to that in Equation 1:

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

This gives us:

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

Now we have the modified system of equations:

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

Next, we can add both equations to eliminate \(x\):

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

This simplifies to:

\[
0x + 2y = 12
\]

So:

\[
2y = 12
\]

Now, divide both sides by 2:

\[
y = 6
\]

Next, we substitute \(y = 6\) back into one of the original equations to find \(x\). Let's use Equation 1:

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

This simplifies to:

\[
4x - 36 = -48
\]

Now, solve for \(x\):

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

The solution to the system of equations is:

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

Therefore, the answer is:

O (-3, 6)