Question

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

Answer 1

To solve the system of equations using the elimination method, we have the following equations:

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

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

First, we notice that the coefficients of \(y\) can be equalized by adjusting Equation 2. We can multiply Equation 2 by 1.5 (which is the factor that would make \(-6y\) in Equation 1 equal to 4y in Equation 2).

Let's calculate that:

Multiply Equation 2 by 1.5:

\[
1.5 \times (-2x + 4y) = 1.5 \times 30
\]

This gives us:

\[
-3x + 6y = 45 \quad \text{(Equation 3)}
\]

Now we have the modified system:

1. \( 4x - 6y = -48 \)
2. \( -3x + 6y = 45 \)

Next, we can add Equation 1 and Equation 3 to eliminate \(y\):

\[
(4x - 6y) + (-3x + 6y) = -48 + 45
\]

This simplifies to:

\[
4x - 3x = -48 + 45
\]

\[
x = -3
\]

Now, substitute \(x = -3\) back into one of the original equations to find \(y\). We'll use Equation 1:

\[
4(-3) - 6y = -48
\]

This simplifies to:

\[
-12 - 6y = -48
\]

Now add 12 to both sides:

\[
-6y = -48 + 12
\]

\[
-6y = -36
\]

Now divide by -6:

\[
y = 6
\]

So the solution to the system of equations is:

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

Thus, the answer is:

**(-3, 6)**.