We are solving the system of three equations:

$-1a - 5b - 1c = -2$ $4a - 5b + 5c = -18$ $-1a - 4b - 4c = 2$

Step 1: Write the equations in standard form

$-1a - 5b - c = -2$ $4a - 5b + 5c = -18$ $-1a - 4b - 4c = 2$

Step 2: Eliminate one variable

To simplify, let's eliminate $a$ from two pairs of equations.

Eliminate $a$ between the first and third equations:

$(-1)(-1a - 5b - c) + (-1a - 4b - 4c) = -(-2) + 2$ $a + 5b + c - a - 4b - 4c = 4$ b - 3c = 4 \tag{4}

Eliminate $a$ between the first and second equations:

$4(-1a - 5b - c) + (-1)(4a - 5b + 5c) = 4(-2) + (-18)$ $-4a - 20b - 4c - 4a + 5b - 5c = -8 - 18$ $-8a - 15b - 9c = -26$ -15b - 9c = -26 \quad \text{(since a terms cancel out)} \tag{5}

Step 3: Solve the two-variable system

We now have two equations: b - 3c = 4 \tag{4} -15b - 9c = -26 \tag{5}

Solve for $b$ in terms of $c$ from Equation (4):

b = 3c + 4 \tag{6}

Substitute Equation (6) into Equation (5):

$-15(3c + 4) - 9c = -26$ $-45c - 60 - 9c = -26$ $-54c - 60 = -26$ $-54c = 34$ c = -\frac{17}{27} \tag{7}

Step 4: Solve for $b$

Substitute $c = -\frac{17}{27}$ into Equation (6): $b = 3\left(-\frac{17}{27}\right) + 4$ $b = -\frac{51}{27} + 4$ $b = -\frac{51}{27} + \frac{108}{27}$ b = \frac{57}{27} = \frac{19}{9} \tag{8}

Step 5: Solve for $a$

Substitute $b = \frac{19}{9}$ and $c = -\frac{17}{27}$ into the first equation: $-1a - 5b - c = -2$ $-1a - 5\left(\frac{19}{9}\right) - \left(-\frac{17}{27}\right) = -2$ $-1a - \frac{95}{9} + \frac{17}{27} = -2$ Convert to a common denominator of 27: $-1a - \frac{285}{27} + \frac{17}{27} = -2$ $-1a - \frac{268}{27} = -2$ $-1a = -2 + \frac{268}{27}$ $-1a = -\frac{54}{27} + \frac{268}{27}$ $-1a = \frac{214}{27}$ a = -\frac{214}{27} \tag{9}

Final Solution:

$a = -\frac{214}{27}, \quad b = \frac{19}{9}, \quad c = -\frac{17}{27}$