You have the system of equations:

1. $x + y - 4 = 0$
2. $2x = 8y - 4$

Let's solve this step-by-step.

Step 1: Rearrange the first equation to express $x$ in terms of $y$:

From $x + y - 4 = 0$, we can isolate $x$: $x = 4 - y$

Step 2: Substitute this expression for $x$ into the second equation:

The second equation is $2x = 8y - 4$. Substitute $x = 4 - y$ into this: $2(4 - y) = 8y - 4$ Simplify: $8 - 2y = 8y - 4$

Step 3: Solve for $y$:

Add $2y$ to both sides: $8 = 10y - 4$ Now, add $4$ to both sides: $12 = 10y$ Finally, divide by 10: $y = \frac{12}{10} = \frac{6}{5}$

Step 4: Substitute $y = \frac{6}{5}$ into the expression for $x$:

Now substitute $y = \frac{6}{5}$ back into $x = 4 - y$: $x = 4 - \frac{6}{5}$ Rewrite 4 as $\frac{20}{5}$: $x = \frac{20}{5} - \frac{6}{5} = \frac{14}{5}$

Final Solution:

The solution to the system of equations is: $x = \frac{14}{5}, \quad y = \frac{6}{5}$

Would you like a more detailed explanation of any step, or do you have other questions on this?

Related Questions:

1. How would the solution change if we had a different coefficient in front of $x$ in the first equation?
2. What is the graphical interpretation of solving this system of linear equations?
3. How can we use substitution in systems of equations involving three variables?
4. Can you solve this system by elimination instead of substitution?
5. How do you check the solution of a system of equations?

Tip: When solving systems of equations, always check the solutions by substituting them back into the original equations to ensure they satisfy both.