Using elimination method, If `2x + 3y = 12` and `3x - 2y = 5` then

Options

A. ` x =2, y = 3 `
B. `x =2, y = - 3 `
C. `x = 3, y = 2 `
D. `x= 3, y = - 2 `

Answer

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

1. \( 2x + 3y = 12 \) (Equation 1)
2. \( 3x - 2y = 5 \) (Equation 2)

### Step 1: Align the equations for elimination
We will manipulate the equations to eliminate one variable. Let's eliminate \( y \). To do this, we need to make the coefficients of \( y \) in both equations equal.

To do this, we can multiply Equation 1 by 2 and Equation 2 by 3:

- Multiply Equation 1 by 2:
\[ 2(2x + 3y) = 2(12) \implies 4x + 6y = 24 \quad (Equation 3) \]

- Multiply Equation 2 by 3:
\[ 3(3x - 2y) = 3(5) \implies 9x - 6y = 15 \quad (Equation 4) \]

### Step 2: Add the equations to eliminate \( y \)
Now, we will add Equation 3 and Equation 4:
\[ (4x + 6y) + (9x - 6y) = 24 + 15 \]
This simplifies to:
\[ 4x + 9x + 6y - 6y = 39 \]
\[ 13x = 39 \]

### Step 3: Solve for \( x \)
Now, divide both sides by 13:
\[ x = \frac{39}{13} = 3 \]

### Step 4: Substitute \( x \) back to find \( y \)
Now that we have \( x = 3 \), we can substitute this value back into either of the original equations to find \( y \). Let's use Equation 1:
\[ 2(3) + 3y = 12 \]
This simplifies to:
\[ 6 + 3y = 12 \]
Subtract 6 from both sides:
\[ 3y = 12 - 6 \]
\[ 3y = 6 \]
Now, divide both sides by 3:
\[ y = \frac{6}{3} = 2 \]

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