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

Answer

To solve the system of linear equations given by:

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

we will use the method of determinants (Cramer's Rule).

### Step 1: Identify the coefficients
We can identify the coefficients from the equations:

- For the first equation \( 5x + 2y = 4 \):
- \( a_1 = 5 \)
- \( b_1 = 2 \)
- \( c_1 = -4 \) (since we rearrange it to \( 5x + 2y - 4 = 0 \))

- For the second equation \( 7x + 3y = 5 \):
- \( a_2 = 7 \)
- \( b_2 = 3 \)
- \( c_2 = -5 \) (similarly rearranging)

### Step 2: Check for a unique solution
To ensure a unique solution exists, we check the condition:

\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

Calculating:

\[ \frac{5}{7} \neq \frac{2}{3} \]

Since this condition holds true, we can conclude that a unique solution exists.

### Step 3: Calculate \( x \) using Cramer's Rule
The formula for \( x \) is given by:

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

Substituting the values:

\[ x = \frac{(2)(-5) - (3)(-4)}{(5)(3) - (7)(2)} \]

Calculating the numerator:

\[ 2 \cdot -5 = -10 \quad \text{and} \quad 3 \cdot -4 = -12 \quad \Rightarrow \quad -10 + 12 = 2 \]

Calculating the denominator:

\[ 5 \cdot 3 = 15 \quad \text{and} \quad 7 \cdot 2 = 14 \quad \Rightarrow \quad 15 - 14 = 1 \]

Thus, we have:

\[ x = \frac{2}{1} = 2 \]

### Step 4: Calculate \( y \) using Cramer's Rule
The formula for \( y \) is given by:

\[ y = \frac{a_2c_1 - a_1c_2}{a_1b_2 - a_2b_1} \]

Substituting the values:

\[ y = \frac{(7)(-4) - (5)(-5)}{(5)(3) - (7)(2)} \]

Calculating the numerator:

\[ 7 \cdot -4 = -28 \quad \text{and} \quad 5 \cdot -5 = -25 \quad \Rightarrow \quad -28 + 25 = -3 \]

The denominator remains the same as before:

\[ 15 - 14 = 1 \]

Thus, we have:

\[ y = \frac{-3}{1} = -3 \]

### Final Solution
The solution to the system of equations is:

\[ x = 2, \quad y = -3 \]