Solve the following system of linear equations: x1 + x2 − 5x3 = 8, -2x1 + 3x2 + x3 = 12, 3x1 + x2 + 4x3 = 5

Multiply equation (1) by 2:

$$2(x_1 + x_2 - 5x_3) = 2(8)$$ $$2x_1 + 2x_2 - 10x_3 = 16 \quad \text{(Equation 4)}$$

Add Equation (4) to Equation (2):

$$(2x_1 + 2x_2 - 10x_3) + (-2x_1 + 3x_2 + x_3) = 16 + 12$$ $$5x_2 - 9x_3 = 28 \quad \text{(Equation 5)}$$

Multiply equation (1) by -3:

$$-3(x_1 + x_2 - 5x_3) = -3(8)$$ $$-3x_1 - 3x_2 + 15x_3 = -24 \quad \text{(Equation 6)}$$

Add Equation (6) to Equation (3):

$$(-3x_1 - 3x_2 + 15x_3) + (3x_1 + x_2 + 4x_3) = -24 + 5$$ $$-2x_2 + 19x_3 = -19 \quad \text{(Equation 7)}$$

The new system is: 5) $5x_2 - 9x_3 = 28$ 7) $-2x_2 + 19x_3 = -19$ Multiply Equation (5) by 2 and Equation (7) by 5:

$$2(5x_2 - 9x_3) = 2(28) \implies 10x_2 - 18x_3 = 56 \quad \text{(Equation 8)}$$ $$5(-2x_2 + 19x_3) = 5(-19) \implies -10x_2 + 95x_3 = -95 \quad \text{(Equation 9)}$$

Add Equation (8) and Equation (9):

$$(10x_2 - 18x_3) + (-10x_2 + 95x_3) = 56 + (-95)$$ $$77x_3 = -39$$ $$x_3 = -\frac{39}{77}$$

$$5x_2 - 9\left(-\frac{39}{77}\right) = 28$$ $$5x_2 + \frac{351}{77} = 28$$ $$5x_2 = 28 - \frac{351}{77}$$ $$5x_2 = \frac{28 \times 77 - 351}{77}$$ $$5x_2 = \frac{2156 - 351}{77}$$ $$5x_2 = \frac{1805}{77}$$ $$x_2 = \frac{1805}{5 \times 77}$$ $$x_2 = \frac{361}{77}$$

$$x_1 + x_2 - 5x_3 = 8$$ $$x_1 + \frac{361}{77} - 5\left(-\frac{39}{77}\right) = 8$$ $$x_1 + \frac{361}{77} + \frac{195}{77} = 8$$ $$x_1 + \frac{361 + 195}{77} = 8$$ $$x_1 + \frac{556}{77} = 8$$ $$x_1 = 8 - \frac{556}{77}$$ $$x_1 = \frac{8 \times 77 - 556}{77}$$ $$x_1 = \frac{616 - 556}{77}$$ $$x_1 = \frac{60}{77}$$

The solution to the system of equations is:

$$x_1 = \frac{60}{77}, \quad x_2 = \frac{361}{77}, \quad x_3 = -\frac{39}{77}$$ $$\boxed{x_1 = \frac{60}{77}, x_2 = \frac{361}{77}, x_3 = -\frac{39}{77}}$$