Solve for x and y:
2x - 3y-13 =0 , 3x -2y+12 =0

Answer

To solve the system of equations:

1. **Equations Given:**
\[ 2x - 3y - 13 = 0 \quad \text{(Equation 1)} \]
\[ 3x - 2y + 12 = 0 \quad \text{(Equation 2)} \]

2. **Rearranging the Equations:**
Rearranging both equations gives:
\[ 2x - 3y = 13 \quad \text{(Equation 1')} \]
\[ 3x - 2y = -12 \quad \text{(Equation 2')} \]

3. **Using the Elimination Method:**
We will eliminate one variable by making the coefficients of \(x\) the same in both equations. Multiply Equation 1' by 3 and Equation 2' by 2:
\[ 3(2x - 3y) = 3(13) \implies 6x - 9y = 39 \quad \text{(Equation 3)} \]
\[ 2(3x - 2y) = 2(-12) \implies 6x - 4y = -24 \quad \text{(Equation 4)} \]

4. **Subtracting the Equations:**
Now, subtract Equation 4 from Equation 3:
\[ (6x - 9y) - (6x - 4y) = 39 - (-24) \]
This simplifies to:
\[ -9y + 4y = 39 + 24 \]
\[ -5y = 63 \]

5. **Solving for \(y\):**
Dividing both sides by -5:
\[ y = \frac{63}{-5} = -\frac{63}{5} \]

6. **Substituting \(y\) back to find \(x\):**
Now substitute \(y\) back into Equation 1':
\[ 2x - 3\left(-\frac{63}{5}\right) = 13 \]
This simplifies to:
\[ 2x + \frac{189}{5} = 13 \]
To eliminate the fraction, multiply the entire equation by 5:
\[ 10x + 189 = 65 \]
Now, isolate \(x\):
\[ 10x = 65 - 189 \]
\[ 10x = -124 \]
\[ x = \frac{-124}{10} = -\frac{62}{5} \]

7. **Final Values:**
Thus, the solution to the system of equations is:
\[ x = -\frac{62}{5}, \quad y = -\frac{63}{5} \]