5x+7y+2=0
4x+6y+3=0

Answer

To solve the system of linear equations given by:

1. \( 5x + 7y + 2 = 0 \)
2. \( 4x + 6y + 3 = 0 \)

we will follow the steps for solving linear equations using the method of determinants (Cramer's Rule).

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in the standard form \( Ax + By + C = 0 \):

1. \( 5x + 7y = -2 \)
2. \( 4x + 6y = -3 \)

### Step 2: Identify coefficients
From the equations, we identify the coefficients:
- For the first equation:
- \( A_1 = 5 \)
- \( B_1 = 7 \)
- \( C_1 = -2 \)

- For the second equation:
- \( A_2 = 4 \)
- \( B_2 = 6 \)
- \( C_2 = -3 \)

### Step 3: Check the condition for a unique solution
The condition for a unique solution is that the ratios \( \frac{A_1}{A_2} \) and \( \frac{B_1}{B_2} \) must not be equal.

Calculating the ratios:
- \( \frac{A_1}{A_2} = \frac{5}{4} \)
- \( \frac{B_1}{B_2} = \frac{7}{6} \)

Since \( \frac{5}{4} \neq \frac{7}{6} \), a unique solution exists.

### Step 4: Use Cramer's Rule to find \( x \) and \( y \)
Using Cramer's Rule, we can find \( x \) and \( y \) as follows:

#### Finding \( x \):
\[ x = \frac{B_1C_2 - B_2C_1}{A_1B_2 - A_2B_1} \]
Substituting the values:
\[ x = \frac{(7)(-3) - (6)(-2)}{(5)(6) - (4)(7)} \]
Calculating the numerator:
\[ = \frac{-21 + 12}{30 - 28} = \frac{-9}{2} \]

#### Finding \( y \):
\[ y = \frac{A_2C_1 - A_1C_2}{A_1B_2 - A_2B_1} \]
Substituting the values:
\[ y = \frac{(4)(-2) - (5)(-3)}{(5)(6) - (4)(7)} \]
Calculating the numerator:
\[ = \frac{-8 + 15}{30 - 28} = \frac{7}{2} \]

### Step 5: Final solution
Thus, the solution to the system of equations is:
\[ x = -\frac{9}{2}, \quad y = \frac{7}{2} \]

### Summary of the solution:
The values of \( x \) and \( y \) are:
\[ \left( -\frac{9}{2}, \frac{7}{2} \right) \]