Solve, using cross - multiplication :
`2x + 3y - 17 = 0 and 3x - 2y - 6 = 0`.

Answer

To solve the equations \(2x + 3y - 17 = 0\) and \(3x - 2y - 6 = 0\) using the cross-multiplication method, we will follow these steps:

### Step 1: Rewrite the equations in standard form
The equations are already in the standard form \(Ax + By + C = 0\):
1. \(2x + 3y - 17 = 0\) (Here, \(a_1 = 2\), \(b_1 = 3\), \(c_1 = -17\))
2. \(3x - 2y - 6 = 0\) (Here, \(a_2 = 3\), \(b_2 = -2\), \(c_2 = -6\))

### Step 2: Set up the cross-multiplication formula
Using the cross-multiplication method, we have:
\[ x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \]
\[ y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \]

### Step 3: Calculate \(x\)
Substituting the values into the formula for \(x\):
\[ x = \frac{3(-6) - (-2)(-17)}{2(-2) - 3(3)} \]
Calculating the numerator:
\[ = \frac{-18 - 34}{-4 - 9} \]
\[ = \frac{-52}{-13} \]
\[ = 4 \]

### Step 4: Calculate \(y\)
Now substituting the values into the formula for \(y\):
\[ y = \frac{-17(3) - (-6)(2)}{2(-2) - 3(3)} \]
Calculating the numerator:
\[ = \frac{-51 + 12}{-4 - 9} \]
\[ = \frac{-39}{-13} \]
\[ = 3 \]

### Step 5: State the solution
Thus, the solution to the system of equations is:
\[ x = 4 \quad \text{and} \quad y = 3 \]