Mister Exam
0.6(x-y)=66,6; 0,7(x+y)=6,3
The solution
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3*(x - y)
--------- = 333/5
5
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
7*(x + y) 63
--------- = --
10 10
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
7*(x + y)/10 = 63/10
Detail solution
Given the system of equations
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
Let's express from equation 1 x
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{3 x}{5} = \left(\frac{3 x}{5} - \frac{3 \left(x - y\right)}{5}\right) + \frac{333}{5}$$
$$\frac{3 x}{5} = \frac{3 y}{5} + \frac{333}{5}$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{3}{5} x}{\frac{3}{5}} = \frac{\frac{3 y}{5} + \frac{333}{5}}{\frac{3}{5}}$$
$$x = y + 111$$
Let's try the obtained element x to 2-th equation
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
We get:
$$\frac{7 \left(y + \left(y + 111\right)\right)}{10} = \frac{63}{10}$$
$$\frac{7 y}{5} + \frac{777}{10} = \frac{63}{10}$$
We move the free summand 777/10 from the left part to the right part performing the sign change
$$\frac{7 y}{5} = - \frac{777}{10} + \frac{63}{10}$$
$$\frac{7 y}{5} = - \frac{357}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{7}{5} y}{\frac{7}{5}} = - \frac{357}{\frac{7}{5} \cdot 5}$$
$$y = -51$$
Because
$$x = y + 111$$
then
$$x = -51 + 111$$
$$x = 60$$
The answer:
$$x = 60$$
$$y = -51$$
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
Let's express from equation 1 x
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{3 x}{5} = \left(\frac{3 x}{5} - \frac{3 \left(x - y\right)}{5}\right) + \frac{333}{5}$$
$$\frac{3 x}{5} = \frac{3 y}{5} + \frac{333}{5}$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{3}{5} x}{\frac{3}{5}} = \frac{\frac{3 y}{5} + \frac{333}{5}}{\frac{3}{5}}$$
$$x = y + 111$$
Let's try the obtained element x to 2-th equation
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
We get:
$$\frac{7 \left(y + \left(y + 111\right)\right)}{10} = \frac{63}{10}$$
$$\frac{7 y}{5} + \frac{777}{10} = \frac{63}{10}$$
We move the free summand 777/10 from the left part to the right part performing the sign change
$$\frac{7 y}{5} = - \frac{777}{10} + \frac{63}{10}$$
$$\frac{7 y}{5} = - \frac{357}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{7}{5} y}{\frac{7}{5}} = - \frac{357}{\frac{7}{5} \cdot 5}$$
$$y = -51$$
Because
$$x = y + 111$$
then
$$x = -51 + 111$$
$$x = 60$$
The answer:
$$x = 60$$
$$y = -51$$
Rapid solution
$$x_{1} = 60$$
=
$$60$$
=
$$y_{1} = -51$$
=
$$-51$$
=
=
$$60$$
=
60
$$y_{1} = -51$$
=
$$-51$$
=
-51
Cramer's rule
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
We give the system of equations to the canonical form
$$\frac{3 x}{5} - \frac{3 y}{5} = \frac{333}{5}$$
$$\frac{7 x}{10} + \frac{7 y}{10} = \frac{63}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{3 x_{1}}{5} - \frac{3 x_{2}}{5}\\\frac{7 x_{1}}{10} + \frac{7 x_{2}}{10}\end{matrix}\right] = \left[\begin{matrix}\frac{333}{5}\\\frac{63}{10}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5}\\\frac{7}{10} & \frac{7}{10}\end{matrix}\right] \right)} = \frac{21}{25}$$
, 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 )
$$x_{1} = \frac{25 \operatorname{det}{\left(\left[\begin{matrix}\frac{333}{5} & - \frac{3}{5}\\\frac{63}{10} & \frac{7}{10}\end{matrix}\right] \right)}}{21} = 60$$
$$x_{2} = \frac{25 \operatorname{det}{\left(\left[\begin{matrix}\frac{3}{5} & \frac{333}{5}\\\frac{7}{10} & \frac{63}{10}\end{matrix}\right] \right)}}{21} = -51$$
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
We give the system of equations to the canonical form
$$\frac{3 x}{5} - \frac{3 y}{5} = \frac{333}{5}$$
$$\frac{7 x}{10} + \frac{7 y}{10} = \frac{63}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{3 x_{1}}{5} - \frac{3 x_{2}}{5}\\\frac{7 x_{1}}{10} + \frac{7 x_{2}}{10}\end{matrix}\right] = \left[\begin{matrix}\frac{333}{5}\\\frac{63}{10}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5}\\\frac{7}{10} & \frac{7}{10}\end{matrix}\right] \right)} = \frac{21}{25}$$
, 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 )
$$x_{1} = \frac{25 \operatorname{det}{\left(\left[\begin{matrix}\frac{333}{5} & - \frac{3}{5}\\\frac{63}{10} & \frac{7}{10}\end{matrix}\right] \right)}}{21} = 60$$
$$x_{2} = \frac{25 \operatorname{det}{\left(\left[\begin{matrix}\frac{3}{5} & \frac{333}{5}\\\frac{7}{10} & \frac{63}{10}\end{matrix}\right] \right)}}{21} = -51$$
Gaussian elimination
Given the system of equations
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
We give the system of equations to the canonical form
$$\frac{3 x}{5} - \frac{3 y}{5} = \frac{333}{5}$$
$$\frac{7 x}{10} + \frac{7 y}{10} = \frac{63}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5} & \frac{333}{5}\\\frac{7}{10} & \frac{7}{10} & \frac{63}{10}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{3}{5}\\\frac{7}{10}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5} & \frac{333}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{7}{10} - \frac{3 \cdot 7}{5 \cdot 6} & \frac{7}{10} - - \frac{7}{10} & \frac{63}{10} - \frac{7 \cdot 333}{5 \cdot 6}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5} & \frac{333}{5}\\0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{3}{5}\\\frac{7}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{3}{5} - \frac{\left(-3\right) 0}{7} & - \frac{3}{5} - \frac{\left(-3\right) 7}{5 \cdot 7} & \frac{333}{5} - - \frac{-153}{5}\end{matrix}\right] = \left[\begin{matrix}\frac{3}{5} & 0 & 36\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{3}{5} & 0 & 36\\0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{3 x_{1}}{5} - 36 = 0$$
$$\frac{7 x_{2}}{5} + \frac{357}{5} = 0$$
We get the answer:
$$x_{1} = 60$$
$$x_{2} = -51$$
$$\frac{3 \left(x - y\right)}{5} = \frac{333}{5}$$
$$\frac{7 \left(x + y\right)}{10} = \frac{63}{10}$$
We give the system of equations to the canonical form
$$\frac{3 x}{5} - \frac{3 y}{5} = \frac{333}{5}$$
$$\frac{7 x}{10} + \frac{7 y}{10} = \frac{63}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5} & \frac{333}{5}\\\frac{7}{10} & \frac{7}{10} & \frac{63}{10}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{3}{5}\\\frac{7}{10}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5} & \frac{333}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{7}{10} - \frac{3 \cdot 7}{5 \cdot 6} & \frac{7}{10} - - \frac{7}{10} & \frac{63}{10} - \frac{7 \cdot 333}{5 \cdot 6}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{3}{5} & - \frac{3}{5} & \frac{333}{5}\\0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{3}{5}\\\frac{7}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{3}{5} - \frac{\left(-3\right) 0}{7} & - \frac{3}{5} - \frac{\left(-3\right) 7}{5 \cdot 7} & \frac{333}{5} - - \frac{-153}{5}\end{matrix}\right] = \left[\begin{matrix}\frac{3}{5} & 0 & 36\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{3}{5} & 0 & 36\\0 & \frac{7}{5} & - \frac{357}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{3 x_{1}}{5} - 36 = 0$$
$$\frac{7 x_{2}}{5} + \frac{357}{5} = 0$$
We get the answer:
$$x_{1} = 60$$
$$x_{2} = -51$$