Solve by cross - multiplication :
3x + y = 13 and x - 3y + 9 = 0.

Answer

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

1. \( 3x + y = 13 \) (Equation 1)
2. \( x - 3y + 9 = 0 \) (Equation 2)

First, we rewrite Equation 2 in a more standard form:

\[ x - 3y = -9 \quad \text{(Rearranging Equation 2)} \]

Now, we have:

1. \( 3x + y = 13 \)
2. \( x - 3y = -9 \)

Next, we will express both equations in the form suitable for cross-multiplication. We can rewrite the equations as follows:

\[ 3x + y - 13 = 0 \quad \text{(Equation 1)} \]
\[ x - 3y + 9 = 0 \quad \text{(Equation 2)} \]

Now, we can set up the cross-multiplication. For two equations of the form \( A_1x + B_1y + C_1 = 0 \) and \( A_2x + B_2y + C_2 = 0 \), the solution can be found using:

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

From our equations, we identify:

- For Equation 1: \( A_1 = 3, B_1 = 1, C_1 = -13 \)
- For Equation 2: \( A_2 = 1, B_2 = -3, C_2 = -9 \)

Now we calculate the denominator:

\[ A_1B_2 - A_2B_1 = 3 \cdot (-3) - 1 \cdot 1 = -9 - 1 = -10 \]

Now we can find \( x \):

\[ x = \frac{B_1C_2 - B_2C_1}{A_1B_2 - A_2B_1} = \frac{1 \cdot (-9) - (-3) \cdot (-13)}{-10} \]
\[ = \frac{-9 - 39}{-10} = \frac{-48}{-10} = \frac{48}{10} = 4.8 \]

Next, we calculate \( y \):

\[ y = \frac{A_2C_1 - A_1C_2}{A_1B_2 - A_2B_1} = \frac{1 \cdot (-13) - 3 \cdot (-9)}{-10} \]
\[ = \frac{-13 + 27}{-10} = \frac{14}{-10} = -1.4 \]

Thus, the solution to the system of equations is:

\[ x = 4.8, \quad y = -1.4 \]