We are given the following system of linear equations:

1. $x - 4y + 2 = 0$
2. $2x + y - 4 = 0$

Step 1: Solve one of the equations for $x$ or $y$

Let's solve the first equation for $x$:

$x - 4y + 2 = 0$ Rearrange it to express $x$ in terms of $y$:

$x = 4y - 2$

Step 2: Substitute into the second equation

Now substitute $x = 4y - 2$ into the second equation $2x + y - 4 = 0$:

$2(4y - 2) + y - 4 = 0$ Simplify the equation:

$8y - 4 + y - 4 = 0$

$9y - 8 = 0$

$9y = 8$

$y = \frac{8}{9}$

Step 3: Find $x$

Substitute $y = \frac{8}{9}$ back into the equation $x = 4y - 2$:

$x = 4\left(\frac{8}{9}\right) - 2$

$x = \frac{32}{9} - 2 = \frac{32}{9} - \frac{18}{9} = \frac{14}{9}$

Final Answer:

$x = \frac{14}{9}, \quad y = \frac{8}{9}$

Would you like a detailed explanation of any of the steps? Here are five related questions you might explore:

1. How do you solve a system of equations using substitution?
2. How would the solution change if we used the elimination method instead?
3. Can this system be solved graphically, and what would the intersection represent?
4. How can we check the solution to a system of linear equations?
5. What happens if a system of linear equations has no solution or infinitely many solutions?

Tip: Substitution is often a good method when one variable is already isolated or easily isolated in one of the equations!