Solve the following system of equations: 2y = x - 3 x + 4y = 1

Okay, let's solve this system of equations step by step.

We have the following system:

$$\begin{cases} 2y = x - 3 \\ x + 4y = 1 \end{cases}$$

First, let's express (x) from the first equation:

$$x = 2y + 3$$

Now, substitute this expression for (x) into the second equation:

$$(2y + 3) + 4y = 1$$

Combine like terms:

$$6y + 3 = 1$$

Subtract 3 from both sides:

$$6y = 1 - 3$$

$$6y = -2$$

Divide by 6:

$$y = \frac{-2}{6} = -\frac{1}{3}$$

Now that we have the value of (y), we can find (x). Substitute (y = -\frac{1}{3}) into the expression for (x):

$$x = 2\left(-\frac{1}{3}\right) + 3$$

$$x = -\frac{2}{3} + 3$$

$$x = -\frac{2}{3} + \frac{9}{3}$$

$$x = \frac{7}{3}$$

So, the solution to the system of equations is:

$$x = \frac{7}{3}, \quad y = -\frac{1}{3}$$