Question

Solve the system of equations below using elimination. You must show all work to earn full credit. 2x+3y=12 4x−3y=6

Answer 1

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

1. \( 2x + 3y = 12 \)
2. \( 4x - 3y = 6 \)

We will eliminate one of the variables by adding or subtracting the equations. In this case, we can eliminate \( y \) by adding the two equations together.

### Step 1: Align the equations
Here are our equations again:
\[
2x + 3y = 12 \quad \quad (1)
\]
\[
4x - 3y = 6 \quad \quad (2)
\]

### Step 2: Add the equations
Adding both equations will eliminate \( y \):

\[
(2x + 3y) + (4x - 3y) = 12 + 6
\]

Combine like terms:

\[
(2x + 4x) + (3y - 3y) = 18
\]
\[
6x + 0y = 18
\]
\[
6x = 18
\]

### Step 3: Solve for \( x \)
Now we solve for \( x \):

\[
x = \frac{18}{6} = 3
\]

### Step 4: Substitute \( x \) back into one of the original equations
We will substitute \( x = 3 \) back into the first equation to find \( y \):

\[
2x + 3y = 12
\]
Substituting \( x \):

\[
2(3) + 3y = 12
\]
\[
6 + 3y = 12
\]

### Step 5: Solve for \( y \)
Now we isolate \( y \):

\[
3y = 12 - 6
\]
\[
3y = 6
\]
\[
y = \frac{6}{3} = 2
\]

### Final result
Thus, the solution to the system of equations is:
\[
x = 3, \quad y = 2
\]

### Verification
Let's verify the solution by plugging \( x = 3 \) and \( y = 2 \) back into both original equations:

1. **First equation:**
\[
2(3) + 3(2) = 6 + 6 = 12 \quad \text{(true)}
\]

2. **Second equation:**
\[
4(3) - 3(2) = 12 - 6 = 6 \quad \text{(true)}
\]

Since both equations are satisfied, the solution is confirmed:

\[
\boxed{(3, 2)}
\]