x=y; 3y=2x+w; 2w=z+y; 2z=2x+y+w; x+y+z+w=1

Given the system of equations
$$x = y$$
$$3 y = w + 2 x$$
$$2 w = y + z$$
$$2 z = w + \left(2 x + y\right)$$
$$w + \left(z + \left(x + y\right)\right) = 1$$

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

We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$- x_{3} = 0$$
$$- x_{1} = 0$$
$$- x_{4} = 0$$
$$- 2 x_{2} = 0$$
$$0 - 1 = 0$$
We get the answer:
This system of equations has no solutions