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Question

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1.
$-a + b + c = 4$
$-a + 2b - 3c = -6$
$2a - 4b + 8c = 18$

Explanation:

Step1: Label the equations

Let:

$$\begin{align} -A + B + C &= 4 \tag{1} \\ -A + 2B - 3C &= -6 \tag{2} \\ 2A - 4B + 8C &= 18 \tag{3} \end{align}$$

Step2: Eliminate A with (2)-(1)

Subtract equation (1) from (2) to remove $A$.

$$ (-A + 2B - 3C) - (-A + B + C) = -6 - 4 $$
$$ B - 4C = -10 \tag{4} $$

Step3: Eliminate A with 2*(1)+(3)

Multiply equation (1) by 2, then add to equation (3).

$$ 2(-A + B + C) + (2A - 4B + 8C) = 2*4 + 18 $$
$$ -2B + 10C = 26 $$

Simplify by dividing by -2:

$$ B - 5C = -13 \tag{5} $$

Step4: Solve for C with (4)-(5)

Subtract equation (5) from equation (4).

$$ (B - 4C) - (B - 5C) = -10 - (-13) $$
$$ C = 3 $$

Step5: Solve for B using (4)

Substitute $C=3$ into equation (4).

$$ B - 4(3) = -10 $$
$$ B - 12 = -10 \implies B = 2 $$

Step6: Solve for A using (1)

Substitute $B=2$, $C=3$ into equation (1).

$$ -A + 2 + 3 = 4 $$
$$ -A + 5 = 4 \implies -A = -1 \implies A = 1 $$

Answer:

$A=1$, $B=2$, $C=3$