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

Multiply equation (2) by 2:

$$2(-2x_1 + 3x_2 + x_3) = 2(12)$$ $$-4x_1 + 6x_2 + 2x_3 = 24 \quad \text{(Equation 4)}$$

Add Equation (1) to Equation (4):

$$(4x_1 + x_2 - 5x_3) + (-4x_1 + 6x_2 + 2x_3) = 8 + 24$$ $$7x_2 - 3x_3 = 32 \quad \text{(Equation 5)}$$

Multiply equation (1) by 3:

$$3(4x_1 + x_2 - 5x_3) = 3(8)$$ $$12x_1 + 3x_2 - 15x_3 = 24 \quad \text{(Equation 6)}$$

Multiply equation (3) by 4:

$$4(3x_1 + x_2 + 4x_3) = 4(5)$$ $$12x_1 + 4x_2 + 16x_3 = 20 \quad \text{(Equation 7)}$$

Subtract Equation (6) from Equation (7):

$$(12x_1 + 4x_2 + 16x_3) - (12x_1 + 3x_2 - 15x_3) = 20 - 24$$ $$12x_1 + 4x_2 + 16x_3 - 12x_1 - 3x_2 + 15x_3 = -4$$ $$x_2 + 31x_3 = -4 \quad \text{(Equation 8)}$$

The new system is: 5) $7x_2 - 3x_3 = 32$ 8) $x_2 + 31x_3 = -4$ From Equation (8), express $x_2$ in terms of $x_3$:

$$x_2 = -4 - 31x_3$$

Substitute this expression for $x_2$ into Equation (5):

$$7(-4 - 31x_3) - 3x_3 = 32$$ $$-28 - 217x_3 - 3x_3 = 32$$ $$-220x_3 = 32 + 28$$ $$-220x_3 = 60$$ $$x_3 = -\frac{60}{220}$$ $$x_3 = -\frac{3}{11}$$

$$x_2 = -4 - 31\left(-\frac{3}{11}\right)$$ $$x_2 = -4 + \frac{93}{11}$$ $$x_2 = \frac{-4 \times 11 + 93}{11}$$ $$x_2 = \frac{-44 + 93}{11}$$ $$x_2 = \frac{49}{11}$$

$$4x_1 + x_2 - 5x_3 = 8$$ $$4x_1 + \frac{49}{11} - 5\left(-\frac{3}{11}\right) = 8$$ $$4x_1 + \frac{49}{11} + \frac{15}{11} = 8$$ $$4x_1 + \frac{49 + 15}{11} = 8$$ $$4x_1 + \frac{64}{11} = 8$$ $$4x_1 = 8 - \frac{64}{11}$$ $$4x_1 = \frac{8 \times 11 - 64}{11}$$ $$4x_1 = \frac{88 - 64}{11}$$ $$4x_1 = \frac{24}{11}$$ $$x_1 = \frac{24}{4 \times 11}$$ $$x_1 = \frac{6}{11}$$

The solution to the system of equations is:

$$x_1 = \frac{6}{11}, \quad x_2 = \frac{49}{11}, \quad x_3 = -\frac{3}{11}$$