Solve the linear equations using cross multiplication method.
`{:(0.4x-1.5y=6.5),(0.3x+0.2y=0.9):}`

Answer

To solve the given linear equations using the cross multiplication method, we will follow these steps:

**Step 1: Write the equations**
The given equations are:
1. \( 0.4x - 1.5y = 6.5 \) (Equation 1)
2. \( 0.3x + 0.2y = 0.9 \) (Equation 2)

**Step 2: Eliminate decimals**
To eliminate the decimals, we can multiply each equation by 10:
1. \( 4x - 15y = 65 \) (Equation 1)
2. \( 3x + 2y = 9 \) (Equation 2)

**Step 3: Rearrange the equations**
We need to rearrange both equations to bring the constant terms to the left side:
1. \( 4x - 15y - 65 = 0 \) (Equation 1)
2. \( 3x + 2y - 9 = 0 \) (Equation 2)

**Step 4: Identify coefficients**
From the equations, we identify the coefficients:
- For Equation 1: \( a_1 = 4, b_1 = -15, c_1 = -65 \)
- For Equation 2: \( a_2 = 3, b_2 = 2, c_2 = -9 \)

**Step 5: Apply the cross multiplication formula**
The cross multiplication formula states:
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]

Calculating each term:
- \( b_1c_2 - b_2c_1 = (-15)(-9) - (2)(-65) = 135 + 130 = 265 \)
- \( c_1a_2 - c_2a_1 = (-65)(3) - (-9)(4) = -195 + 36 = -159 \)
- \( a_1b_2 - a_2b_1 = (4)(2) - (3)(-15) = 8 + 45 = 53 \)

**Step 6: Set up the equations**
Now we can set up the equations based on the cross multiplication:
\[ \frac{x}{265} = \frac{y}{-159} = \frac{1}{53} \]

**Step 7: Solve for x and y**
From \( \frac{x}{265} = \frac{1}{53} \):
\[ x = \frac{265}{53} \implies x = 5 \]

From \( \frac{y}{-159} = \frac{1}{53} \):
\[ y = \frac{-159}{53} \implies y = -3 \]

**Final Solution:**
The solution to the system of equations is:
\[ x = 5, \quad y = -3 \]