3x+4y-5=0
x-y+3=0

Answer

To solve the system of linear equations given by:

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

using the matrix method, we will follow these steps:

### Step 1: Rewrite the equations in standard form
We need to express both equations in the form \( ax + by + c = 0 \).

- The first equation is already in standard form: \( 3x + 4y - 5 = 0 \).
- The second equation can be rewritten as \( x - y + 3 = 0 \).

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \( 3x + 4y - 5 = 0 \):
- \( a_1 = 3 \), \( b_1 = 4 \), \( c_1 = -5 \)

- For the second equation \( x - y + 3 = 0 \):
- \( a_2 = 1 \), \( b_2 = -1 \), \( c_2 = -3 \)

### Step 3: Calculate the determinants
We will calculate the determinants \( \Delta \), \( \Delta_1 \), and \( \Delta_2 \).

1. **Determinant \( \Delta \)**:
\[ \Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = \begin{vmatrix} 3 & 4 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (4)(1) = -3 - 4 = -7 \]

2. **Determinant \( \Delta_1 \)**:
\[ \Delta_1 = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = \begin{vmatrix} -5 & 4 \\ -3 & -1 \end{vmatrix} = (-5)(-1) - (4)(-3) = 5 + 12 = 17 \]

3. **Determinant \( \Delta_2 \)**:
\[ \Delta_2 = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = \begin{vmatrix} 3 & -5 \\ 1 & -3 \end{vmatrix} = (3)(-3) - (-5)(1) = -9 + 5 = -4 \]

### Step 4: Calculate the values of \( x \) and \( y \)
Using the determinants, we can find \( x \) and \( y \):

1. **Calculate \( x \)**:
\[ x = \frac{\Delta_1}{\Delta} = \frac{17}{-7} = -\frac{17}{7} \]

2. **Calculate \( y \)**:
\[ y = \frac{\Delta_2}{\Delta} = \frac{-4}{-7} = \frac{4}{7} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = -\frac{17}{7}, \quad y = \frac{4}{7} \]