Mister Exam
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How to use it?
- 2x-2y+2=0; 2x-2y-2=0
- -3x+5y=9; 11x-3y=-13
-
Х+1=2у; 5ху+у^2-16=0
- 4х+2у=5; 4х-6у=-7
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Identical expressions
- two x- two y+2= zero ; 2x-2y-2= zero
- 2x minus 2y plus 2 equally 0; 2x minus 2y minus 2 equally 0
- two x minus two y plus 2 equally zero ; 2x minus 2y minus 2 equally zero
- 2x-2y+2=O; 2x-2y-2=O
-
Similar expressions
- 2x-2y+2=0; 2x+2y-2=0
- 2x-2y+2=0; 2x-2y+2=0
- 2x+2y+2=0; 2x-2y-2=0
- 2x-2y-2=0; 2x-2y-2=0
2x-2y+2=0; 2x-2y-2=0
The solution
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[src]
2*x - 2*y + 2 = 0
$$\left(2 x - 2 y\right) + 2 = 0$$
2*x - 2*y - 2 = 0
$$\left(2 x - 2 y\right) - 2 = 0$$
2*x - 2*y - 2 = 0
Detail solution
Given the system of equations
$$\left(2 x - 2 y\right) + 2 = 0$$
$$\left(2 x - 2 y\right) - 2 = 0$$
Let's express from equation 1 x
$$\left(2 x - 2 y\right) + 2 = 0$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x + 2 = 2 y$$
$$2 x + 2 = 2 y$$
We move the free summand 2 from the left part to the right part performing the sign change
$$2 x = 2 y - 2$$
$$2 x = 2 y - 2$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{2 y - 2}{2}$$
$$x = y - 1$$
Let's try the obtained element x to 2-th equation
$$\left(2 x - 2 y\right) - 2 = 0$$
We get:
$$\left(- 2 y + 2 \left(y - 1\right)\right) - 2 = 0$$
so
This system of equations has no solutions
$$\left(2 x - 2 y\right) + 2 = 0$$
$$\left(2 x - 2 y\right) - 2 = 0$$
Let's express from equation 1 x
$$\left(2 x - 2 y\right) + 2 = 0$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x + 2 = 2 y$$
$$2 x + 2 = 2 y$$
We move the free summand 2 from the left part to the right part performing the sign change
$$2 x = 2 y - 2$$
$$2 x = 2 y - 2$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{2 y - 2}{2}$$
$$x = y - 1$$
Let's try the obtained element x to 2-th equation
$$\left(2 x - 2 y\right) - 2 = 0$$
We get:
$$\left(- 2 y + 2 \left(y - 1\right)\right) - 2 = 0$$
so
This system of equations has no solutions
Gaussian elimination
Given the system of equations
$$\left(2 x - 2 y\right) + 2 = 0$$
$$\left(2 x - 2 y\right) - 2 = 0$$
We give the system of equations to the canonical form
$$2 x - 2 y = -2$$
$$2 x - 2 y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -2 & -2\\2 & -2 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -2 & -2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & -2 - -2 & 2 - -2\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -2 & -2\\0 & 0 & 4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -2 & -2\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$2 x_{1} - 2 x_{2} + 2 = 0$$
$$0 - 4 = 0$$
We get the answer:
This system of equations has no solutions
$$\left(2 x - 2 y\right) + 2 = 0$$
$$\left(2 x - 2 y\right) - 2 = 0$$
We give the system of equations to the canonical form
$$2 x - 2 y = -2$$
$$2 x - 2 y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -2 & -2\\2 & -2 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -2 & -2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & -2 - -2 & 2 - -2\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -2 & -2\\0 & 0 & 4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -2 & -2\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$2 x_{1} - 2 x_{2} + 2 = 0$$
$$0 - 4 = 0$$
We get the answer:
This system of equations has no solutions