Question

\begin{aligned}
&-x+4y=4
\\\\
&-x+3y=1
\end{aligned}

Answer 1

To solve the system of equations

\[
\begin{aligned}
1) & \quad -x + 4y = 4 \\
2) & \quad -x + 3y = 1
\end{aligned}
\]

we can use the method of elimination or substitution. Here, I will use elimination.

First, let's start by rewriting both equations in a more standard form (isolating \(x\) in both cases):

From equation 1, we can express \(x\):

\[
-x = 4 - 4y \implies x = 4y - 4 \tag{3}
\]

From equation 2, we can also express \(x\):

\[
-x = 1 - 3y \implies x = 3y - 1 \tag{4}
\]

Now we have two expressions for \(x\):

\[
4y - 4 = 3y - 1
\]

Next, we can solve for \(y\). Subtract \(3y\) from both sides:

\[
4y - 3y - 4 = -1
\]

This simplifies to:

\[
y - 4 = -1
\]

Adding 4 to both sides gives:

\[
y = 3
\]

Now that we have \(y\), we can substitute it back into either equation (I will use equation 3):

\[
x = 4(3) - 4 = 12 - 4 = 8
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (8, 3)
\]

To summarize, the final answer is:

\[
\boxed{(8, 3)}
\]