5х-7у=9; 6х+5у=-16

Given the system of equations
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$

We give the system of equations to the canonical form
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -7 & 9\\6 & 5 & -16\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -7 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{5 \cdot 6}{5} & 5 - - \frac{42}{5} & -16 - \frac{6 \cdot 9}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -7 & 9\\0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\\frac{67}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-35\right) 0}{67} & -7 - \frac{\left(-35\right) 67}{5 \cdot 67} & 9 - - -14\end{matrix}\right] = \left[\begin{matrix}5 & 0 & -5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5\\0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$

It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} + 5 = 0$$
$$\frac{67 x_{2}}{5} + \frac{134}{5} = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = -2$$