Solve, using cross - multiplication :
4x + 6y = 15
3x - 4y = 7

Answer

To solve the system of equations using the cross-multiplication method, we have the following equations:

1. \( 4x + 6y = 15 \)
2. \( 3x - 4y = 7 \)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in the form \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \).

From the first equation:
\[ 4x + 6y - 15 = 0 \quad \Rightarrow \quad a_1 = 4, \, b_1 = 6, \, c_1 = -15 \]

From the second equation:
\[ 3x - 4y - 7 = 0 \quad \Rightarrow \quad a_2 = 3, \, b_2 = -4, \, c_2 = -7 \]

### Step 2: Set up the cross-multiplication formulas
Using the cross-multiplication method, we can express \( x \) and \( y \) as follows:

\[ x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \]
\[ y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \]

### Step 3: Calculate \( x \)
Substituting the values into the formula for \( x \):

\[ x = \frac{(6)(-7) - (-4)(-15)}{(4)(-4) - (3)(6)} \]

Calculating the numerator:
\[ 6 \cdot -7 = -42 \]
\[ -4 \cdot -15 = 60 \]
So,
\[ b_1c_2 - b_2c_1 = -42 - 60 = -102 \]

Calculating the denominator:
\[ 4 \cdot -4 = -16 \]
\[ 3 \cdot 6 = 18 \]
So,
\[ a_1b_2 - a_2b_1 = -16 - 18 = -34 \]

Thus, we have:
\[ x = \frac{-102}{-34} = 3 \]

### Step 4: Calculate \( y \)
Now substituting the values into the formula for \( y \):

\[ y = \frac{(-15)(3) - (-7)(4)}{(4)(-4) - (3)(6)} \]

Calculating the numerator:
\[ -15 \cdot 3 = -45 \]
\[ -7 \cdot 4 = -28 \]
So,
\[ c_1a_2 - c_2a_1 = -45 + 28 = -17 \]

Using the same denominator as before:
\[ a_1b_2 - a_2b_1 = -34 \]

Thus, we have:
\[ y = \frac{-17}{-34} = \frac{1}{2} \]

### Final Solution
The solution to the system of equations is:
\[ x = 3, \quad y = \frac{1}{2} \]

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