Solve 3x + 2y = 11 and 2x + 3y = 4.

Options

A. x = 5, y = -2
B. x = 5, y = 2
C. x = -5, y = -2
D. none of these

Answer

To solve the system of equations given by:

1. \( 3x + 2y = 11 \) (Equation 1)
2. \( 2x + 3y = 4 \) (Equation 2)

We will use the elimination method.

### Step 1: Make the coefficients of \( x \) the same
We can multiply Equation 1 by 2 and Equation 2 by 3 to make the coefficients of \( x \) the same.

- Multiply Equation 1 by 2:
\[ 2(3x + 2y) = 2(11) \implies 6x + 4y = 22 \quad \text{(Equation 3)} \]

- Multiply Equation 2 by 3:
\[ 3(2x + 3y) = 3(4) \implies 6x + 9y = 12 \quad \text{(Equation 4)} \]

### Step 2: Subtract the two new equations
Now, we will subtract Equation 3 from Equation 4 to eliminate \( x \):

\[ (6x + 9y) - (6x + 4y) = 12 - 22 \]

This simplifies to:
\[ 9y - 4y = -10 \]
\[ 5y = -10 \]

### Step 3: Solve for \( y \)
Now, divide both sides by 5:
\[ y = \frac{-10}{5} = -2 \]

### Step 4: Substitute \( y \) back into one of the original equations
Now that we have \( y \), we can substitute it back into Equation 1 to find \( x \):

\[ 3x + 2(-2) = 11 \]
\[ 3x - 4 = 11 \]

### Step 5: Solve for \( x \)
Add 4 to both sides:
\[ 3x = 11 + 4 \]
\[ 3x = 15 \]

Now, divide by 3:
\[ x = \frac{15}{3} = 5 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 5, \quad y = -2 \]