Solve by cross - multiplication :
`2x + 3y = 6 and 6x - 5y = 4`.

Answer

To solve the given simultaneous equations using the cross-multiplication method, we have the equations:

1. \( 2x + 3y = 6 \) (Equation 1)
2. \( 6x - 5y = 4 \) (Equation 2)

### Step 1: Write the equations in standard form
The equations are already in standard form, which is \( Ax + By = C \).

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1: \( A_1 = 2, B_1 = 3, C_1 = 6 \)
- For Equation 2: \( A_2 = 6, B_2 = -5, C_2 = 4 \)

### Step 3: Set up the cross-multiplication
Using the cross-multiplication method, we set up the following ratios:

\[ \frac{x}{B_1C_2 - B_2C_1} = \frac{y}{A_2C_1 - A_1C_2} = \frac{1}{A_1B_2 - A_2B_1} \]

### Step 4: Calculate the determinants
1. Calculate \( B_1C_2 - B_2C_1 \):
\[ B_1C_2 - B_2C_1 = 3 \cdot 4 - (-5) \cdot 6 = 12 + 30 = 42 \]

2. Calculate \( A_2C_1 - A_1C_2 \):
\[ A_2C_1 - A_1C_2 = 6 \cdot 6 - 2 \cdot 4 = 36 - 8 = 28 \]

3. Calculate \( A_1B_2 - A_2B_1 \):
\[ A_1B_2 - A_2B_1 = 2 \cdot (-5) - 6 \cdot 3 = -10 - 18 = -28 \]

### Step 5: Substitute into the ratios
Now we have:
\[ \frac{x}{42} = \frac{y}{28} = \frac{1}{-28} \]

### Step 6: Solve for \( x \) and \( y \)
From the first ratio:
\[ x = 42 \cdot \frac{1}{-28} = -\frac{42}{28} = -\frac{3}{2} \]

From the second ratio:
\[ y = 28 \cdot \frac{1}{-28} = -1 \]

### Final Result
Thus, the solution to the simultaneous equations is:
\[ x = -\frac{3}{2}, \quad y = -1 \]