Question

Solve using the elimination method.

3x+3y=12
6x-3y=6

Answer 1

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

1. \( 3x + 3y = 12 \) (Equation 1)
2. \( 6x - 3y = 6 \) (Equation 2)

First, we can manipulate the equations to easily eliminate one of the variables. In this case, we can eliminate \( y \) by adding the two equations together.

But first, let's rewrite Equation 1 to make it simpler:

\[
3x + 3y = 12 \implies x + y = 4 \quad \text{(Divide the entire equation by 3)}
\]
So, we have:
1. \( x + y = 4 \) (Equation 1)
2. \( 6x - 3y = 6 \) (Equation 2)

Next, we can multiply Equation 1 by 3 to align the coefficients of \( y \):

\[
3(x + y) = 3(4) \implies 3x + 3y = 12 \quad \text{(This is actually just the original Equation 1)}
\]
Let's multiply Equation 1 by 1 (no change), and rewrite Equation 2 with respect to \( y \):

From Equation 1, we have:
\[
y = 4 - x
\]

Substituting \( y \) in Equation 2 gives us:
\[
6x - 3(4 - x) = 6
\]

Expanding this:
\[
6x - 12 + 3x = 6
\]
Combining like terms:
\[
9x - 12 = 6
\]
Adding \( 12 \) to both sides:
\[
9x = 18
\]
Dividing by \( 9 \):
\[
x = 2
\]

Now that we have \( x = 2 \), we substitute this value back into Equation 1 to find \( y \):
\[
x + y = 4 \implies 2 + y = 4
\]
Subtracting \( 2 \) from both sides:
\[
y = 2
\]

Thus, the solution to the system of equations is:
\[
\boxed{(2, 2)}
\]