Mister Exam
x-2y-5=0; 2x-4y+3=0
The solution
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x - 2*y - 5 = 0
$$\left(x - 2 y\right) - 5 = 0$$
2*x - 4*y + 3 = 0
$$\left(2 x - 4 y\right) + 3 = 0$$
2*x - 4*y + 3 = 0
Detail solution
Given the system of equations
$$\left(x - 2 y\right) - 5 = 0$$
$$\left(2 x - 4 y\right) + 3 = 0$$
Let's express from equation 1 x
$$\left(x - 2 y\right) - 5 = 0$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x - 5 = 2 y$$
$$x - 5 = 2 y$$
We move the free summand -5 from the left part to the right part performing the sign change
$$x = 2 y + 5$$
$$x = 2 y + 5$$
Let's try the obtained element x to 2-th equation
$$\left(2 x - 4 y\right) + 3 = 0$$
We get:
$$\left(- 4 y + 2 \left(2 y + 5\right)\right) + 3 = 0$$
so
This system of equations has no solutions
$$\left(x - 2 y\right) - 5 = 0$$
$$\left(2 x - 4 y\right) + 3 = 0$$
Let's express from equation 1 x
$$\left(x - 2 y\right) - 5 = 0$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x - 5 = 2 y$$
$$x - 5 = 2 y$$
We move the free summand -5 from the left part to the right part performing the sign change
$$x = 2 y + 5$$
$$x = 2 y + 5$$
Let's try the obtained element x to 2-th equation
$$\left(2 x - 4 y\right) + 3 = 0$$
We get:
$$\left(- 4 y + 2 \left(2 y + 5\right)\right) + 3 = 0$$
so
This system of equations has no solutions
Gaussian elimination
Given the system of equations
$$\left(x - 2 y\right) - 5 = 0$$
$$\left(2 x - 4 y\right) + 3 = 0$$
We give the system of equations to the canonical form
$$x - 2 y = 5$$
$$2 x - 4 y = -3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -2 & 5\\2 & -4 & -3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -2 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & -4 - - 4 & - 2 \cdot 5 - 3\end{matrix}\right] = \left[\begin{matrix}0 & 0 & -13\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -2 & 5\\0 & 0 & -13\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -2 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$x_{1} - 2 x_{2} - 5 = 0$$
$$0 + 13 = 0$$
We get the answer:
This system of equations has no solutions
$$\left(x - 2 y\right) - 5 = 0$$
$$\left(2 x - 4 y\right) + 3 = 0$$
We give the system of equations to the canonical form
$$x - 2 y = 5$$
$$2 x - 4 y = -3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -2 & 5\\2 & -4 & -3\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -2 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & -4 - - 4 & - 2 \cdot 5 - 3\end{matrix}\right] = \left[\begin{matrix}0 & 0 & -13\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -2 & 5\\0 & 0 & -13\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -2 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$x_{1} - 2 x_{2} - 5 = 0$$
$$0 + 13 = 0$$
We get the answer:
This system of equations has no solutions