Question

Solve the system of two equations using elimination with multiplications. 4x - 6y = -48 -2x + 4y = 30

Answer 1

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

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

To eliminate one of the variables, we can manipulate the equations. Let's eliminate \( x \) by making the coefficients of \( x \) equal.

First, we can multiply the second equation by 2 to match the coefficient of \( x \) in the first equation:

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

This gives us:

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

Now we have the system:

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

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

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

This simplifies to:

\[
0x + 2y = 12
\]

Which simplifies further to:

\[
2y = 12
\]

Now, divide both sides by 2:

\[
y = 6
\]

Now that we have \( y \), we can substitute \( y \) back into one of the original equations to solve for \( x \). We will use the second equation:

\[
-2x + 4(6) = 30
\]

This simplifies to:

\[
-2x + 24 = 30
\]

Subtract 24 from both sides:

\[
-2x = 30 - 24
\]

\[
-2x = 6
\]

Now divide by -2:

\[
x = -3
\]

The solution to the system of equations is:

\[
\boxed{(-3, 6)}
\]