Solve the System of Linear Equations: 6x - 2y = 24 and x + 5y = 4

To solve this problem, we'll begin with the system of equations:

$\begin{aligned} & 6x - 2y = 24 \quad \text{(Equation 1)} \\ & x + 5y = 4 \quad \text{(Equation 2)} \end{aligned}$

We will use the elimination method to solve for $x$ and $y$. First, let's align the equations to eliminate $y$.

We notice that if we multiply Equation 2 by 2, it becomes easier to align coefficients with Equation 1:

$2(x + 5y) = 2 \times 4$

Simplifying gives:

$2x + 10y = 8 \quad \text{(Equation 3)}$

Now, we have:

$\begin{aligned} & 6x - 2y = 24 \\ & 2x + 10y = 8 \\ \end{aligned}$

To eliminate $y$, let's add the equations after aligning coefficients. Multiply Equation 1 by 5 and Equation 3 by 1 to eliminate $y$:

$\begin{aligned} 5(6x - 2y) &= 5(24) \\ 1(2x + 10y) &= 1(8) \\ \end{aligned}$

Which gives:

$\begin{aligned} 30x - 10y &= 120 \\ 2x + 10y &= 8 \end{aligned}$

Adding these:

$(30x - 10y) + (2x + 10y) = 120 + 8$

$32x = 128$

Solving for $x$, we divide by 32:

$x = \frac{128}{32} = 4$

Substitute $x = 4$ back into Equation 2:

$4 + 5y = 4$

Subtract 4 on both sides:

$5y = 0$

Dividing by 5 gives:

$y = 0$

Thus, we have determined the solution to the system of equations:

$x = 4, y = 0$.