We are given a system of four linear equations with four variables $a$, $b$, $c$, and $d$: \begin{align*} \label{eq:1} a + b + c + d &= 4 \\ 3a + 2b + c &= 0 \\ 27a + 9b + 3c + d &= 0 \\ 27a + 6b + c &= 0\end{align*} We need to solve this system of equations for $a, b, c, d$.

1. Problem Description

We are given a system of four linear equations with four variables $a$, $b$, $c$, and $d$:

\begin{align*} \label{eq:1} a + b + c + d &= 4 \\ 3a + 2b + c &= 0 \\ 27a + 9b + 3c + d &= 0 \\ 27a + 6b + c &= 0\end{align*}

We need to solve this system of equations for $a, b, c, d$.

2. Solution Steps

First, we subtract equation (1) from equation (3):

$(27a + 9b + 3c + d) - (a + b + c + d) = 0 - 4$

$26a + 8b + 2c = -4$

$13a + 4b + c = -2$ (5)

Next, we subtract equation (2) from equation (5):

$(13a + 4b + c) - (3a + 2b + c) = -2 - 0$

$10a + 2b = -2$

$5a + b = -1$

$b = -1 - 5a$ (6)

Now, substitute equation (6) into equation (4):

$27a + 6(-1 - 5a) + c = 0$

$27a - 6 - 30a + c = 0$

$-3a - 6 + c = 0$

$c = 3a + 6$ (7)

Substitute equation (6) and (7) into equation (2):

$3a + 2(-1 - 5a) + (3a + 6) = 0$

$3a - 2 - 10a + 3a + 6 = 0$

$-4a + 4 = 0$

$4a = 4$

$a = 1$

Substitute $a = 1$ into equation (6):

$b = -1 - 5(1) = -1 - 5 = -6$

Substitute $a = 1$ into equation (7):

$c = 3(1) + 6 = 3 + 6 = 9$

Substitute $a = 1, b = -6, c = 9$ into equation (1):

$1 + (-6) + 9 + d = 4$

$1 - 6 + 9 + d = 4$

$4 + d = 4$

$d = 0$

3. Final Answer

The solution is $a = 1, b = -6, c = 9, d = 0$.