Solve, using cross - multiplication :
0.4x + 1.5y = 6.5
0.3x + 0.2y = 0.9

Answer

To solve the system of equations using cross multiplication, we start with the given equations:

1. \( 0.4x + 1.5y = 6.5 \) (Equation 1)
2. \( 0.3x + 0.2y = 0.9 \) (Equation 2)

### Step 1: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = 0.4 \)
- \( b_1 = 1.5 \)
- \( c_1 = 6.5 \)

- For Equation 2:
- \( a_2 = 0.3 \)
- \( b_2 = 0.2 \)
- \( c_2 = 0.9 \)

### Step 2: Apply the cross multiplication formula
The formulas for \( x \) and \( y \) using cross multiplication are:
\[ x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \]
\[ y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \]

### Step 3: Calculate \( x \)
Substituting the values into the formula for \( x \):
\[ x = \frac{(1.5 \cdot 0.9) - (0.2 \cdot 6.5)}{(0.4 \cdot 0.2) - (0.3 \cdot 1.5)} \]

Calculating the numerator:
\[ 1.5 \cdot 0.9 = 1.35 \]
\[ 0.2 \cdot 6.5 = 1.3 \]
So,
\[ 1.35 - 1.3 = 0.05 \]

Calculating the denominator:
\[ 0.4 \cdot 0.2 = 0.08 \]
\[ 0.3 \cdot 1.5 = 0.45 \]
So,
\[ 0.08 - 0.45 = -0.37 \]

Now substituting back:
\[ x = \frac{0.05}{-0.37} \approx -0.135135135 \]
Thus,
\[ x \approx -\frac{5}{37} \]

### Step 4: Calculate \( y \)
Now substituting the values into the formula for \( y \):
\[ y = \frac{(6.5 \cdot 0.3) - (0.9 \cdot 0.4)}{(0.4 \cdot 0.2) - (0.3 \cdot 1.5)} \]

Calculating the numerator:
\[ 6.5 \cdot 0.3 = 1.95 \]
\[ 0.9 \cdot 0.4 = 0.36 \]
So,
\[ 1.95 - 0.36 = 1.59 \]

Using the same denominator as before:
\[ 0.08 - 0.45 = -0.37 \]

Now substituting back:
\[ y = \frac{1.59}{-0.37} \approx -4.297297297 \]
Thus,
\[ y \approx -\frac{159}{37} \]

### Final Answers
The solutions to the equations are:
\[ x \approx -\frac{5}{37}, \quad y \approx -\frac{159}{37} \]