4x+у=3; 6x-5y=11

Given the system of equations
$$4 x + y = 3$$
$$6 x - 5 y = 11$$

We give the system of equations to the canonical form
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 1 & 3\\6 & -5 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 1 & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & -5 + \frac{\left(-1\right) 3}{2} & 11 - \frac{3 \cdot 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 1 & 3\\0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{13}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-2\right) 0}{13} & 1 - - -1 & 3 - \frac{\left(-2\right) 13}{2 \cdot 13}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 4\\0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$

It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 4 = 0$$
$$- \frac{13 x_{2}}{2} - \frac{13}{2} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -1$$