Mister Exam
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- Complex numbers step by step
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How to use it?
- 6х-7y=2; 5x-6y=1
- (x+y+2x)(3x-2y); (x+y²+z)(x-z)
-
16x+32y-212,479=0; 32x+128y-791,249=0
- X+y-z=1; 8x+3y-6z=-1; 4x-y+3z=2
-
Identical expressions
- 6х-7y= two ; 5x-6y= one
- 6х minus 7y equally 2; 5x minus 6y equally 1
- 6х minus 7y equally two ; 5x minus 6y equally one
-
Similar expressions
- 6х-7y=2; 5x+6y=1
- 6х+7y=2; 5x-6y=1
6х-7y=2; 5x-6y=1
The solution
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6*x - 7*y = 2
$$6 x - 7 y = 2$$
5*x - 6*y = 1
$$5 x - 6 y = 1$$
5*x - 6*y = 1
Detail solution
Given the system of equations
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
Let's express from equation 1 x
$$6 x - 7 y = 2$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$6 x = 7 y + 2$$
$$6 x = 7 y + 2$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{6 x}{6} = \frac{7 y + 2}{6}$$
$$x = \frac{7 y}{6} + \frac{1}{3}$$
Let's try the obtained element x to 2-th equation
$$5 x - 6 y = 1$$
We get:
$$- 6 y + 5 \left(\frac{7 y}{6} + \frac{1}{3}\right) = 1$$
$$\frac{5}{3} - \frac{y}{6} = 1$$
We move the free summand 5/3 from the left part to the right part performing the sign change
$$- \frac{y}{6} = - \frac{5}{3} + 1$$
$$- \frac{y}{6} = - \frac{2}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{1}{6} y}{- \frac{1}{6}} = - \frac{2}{\left(- \frac{1}{6}\right) 3}$$
$$y = 4$$
Because
$$x = \frac{7 y}{6} + \frac{1}{3}$$
then
$$x = \frac{1}{3} + \frac{4 \cdot 7}{6}$$
$$x = 5$$
The answer:
$$x = 5$$
$$y = 4$$
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
Let's express from equation 1 x
$$6 x - 7 y = 2$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$6 x = 7 y + 2$$
$$6 x = 7 y + 2$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{6 x}{6} = \frac{7 y + 2}{6}$$
$$x = \frac{7 y}{6} + \frac{1}{3}$$
Let's try the obtained element x to 2-th equation
$$5 x - 6 y = 1$$
We get:
$$- 6 y + 5 \left(\frac{7 y}{6} + \frac{1}{3}\right) = 1$$
$$\frac{5}{3} - \frac{y}{6} = 1$$
We move the free summand 5/3 from the left part to the right part performing the sign change
$$- \frac{y}{6} = - \frac{5}{3} + 1$$
$$- \frac{y}{6} = - \frac{2}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{1}{6} y}{- \frac{1}{6}} = - \frac{2}{\left(- \frac{1}{6}\right) 3}$$
$$y = 4$$
Because
$$x = \frac{7 y}{6} + \frac{1}{3}$$
then
$$x = \frac{1}{3} + \frac{4 \cdot 7}{6}$$
$$x = 5$$
The answer:
$$x = 5$$
$$y = 4$$
Rapid solution
$$x_{1} = 5$$
=
$$5$$
=
$$y_{1} = 4$$
=
$$4$$
=
=
$$5$$
=
5
$$y_{1} = 4$$
=
$$4$$
=
4
Cramer's rule
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
We give the system of equations to the canonical form
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 x_{1} - 7 x_{2}\\5 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}2\\1\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}6 & -7\\5 & -6\end{matrix}\right] \right)} = -1$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = - \operatorname{det}{\left(\left[\begin{matrix}2 & -7\\1 & -6\end{matrix}\right] \right)} = 5$$
$$x_{2} = - \operatorname{det}{\left(\left[\begin{matrix}6 & 2\\5 & 1\end{matrix}\right] \right)} = 4$$
$$5 x - 6 y = 1$$
We give the system of equations to the canonical form
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 x_{1} - 7 x_{2}\\5 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}2\\1\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}6 & -7\\5 & -6\end{matrix}\right] \right)} = -1$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = - \operatorname{det}{\left(\left[\begin{matrix}2 & -7\\1 & -6\end{matrix}\right] \right)} = 5$$
$$x_{2} = - \operatorname{det}{\left(\left[\begin{matrix}6 & 2\\5 & 1\end{matrix}\right] \right)} = 4$$
Gaussian elimination
Given the system of equations
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
We give the system of equations to the canonical form
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 & -7 & 2\\5 & -6 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}6\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}6 & -7 & 2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{5 \cdot 6}{6} & -6 - - \frac{35}{6} & 1 - \frac{2 \cdot 5}{6}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & -7 & 2\\0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\- \frac{1}{6}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 0 \cdot 42 & -7 - \frac{\left(-1\right) 42}{6} & 2 - \frac{\left(-2\right) 42}{3}\end{matrix}\right] = \left[\begin{matrix}6 & 0 & 30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & 0 & 30\\0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$6 x_{1} - 30 = 0$$
$$\frac{2}{3} - \frac{x_{2}}{6} = 0$$
We get the answer:
$$x_{1} = 5$$
$$x_{2} = 4$$
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
We give the system of equations to the canonical form
$$6 x - 7 y = 2$$
$$5 x - 6 y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}6 & -7 & 2\\5 & -6 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}6\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}6 & -7 & 2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{5 \cdot 6}{6} & -6 - - \frac{35}{6} & 1 - \frac{2 \cdot 5}{6}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & -7 & 2\\0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\- \frac{1}{6}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 0 \cdot 42 & -7 - \frac{\left(-1\right) 42}{6} & 2 - \frac{\left(-2\right) 42}{3}\end{matrix}\right] = \left[\begin{matrix}6 & 0 & 30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}6 & 0 & 30\\0 & - \frac{1}{6} & - \frac{2}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$6 x_{1} - 30 = 0$$
$$\frac{2}{3} - \frac{x_{2}}{6} = 0$$
We get the answer:
$$x_{1} = 5$$
$$x_{2} = 4$$