Mister Exam
2-5×(0,2y-2x)=3×(3x+2)+2y; 4×(x-2y)-(2x+y)=2-2×(2x+y)
The solution
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/y \
2 - 5*|- - 2*x| = 3*(3*x + 2) + 2*y
\5 /
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
4*(x - 2*y) + -2*x - y = 2 - 2*(2*x + y)
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
-2*x - y + 4*(x - 2*y) = 2 - 2*(2*x + y)
Detail solution
Given the system of equations
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
Let's express from equation 1 x
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
Let's move the summand with the variable x from the right part to the left part performing the sign change
$$\left(2 - 5 \left(- 2 x + \frac{y}{5}\right)\right) + \left(6 - 3 \left(3 x + 2\right)\right) = 2 y + 6$$
$$x - y + 2 = 2 y + 6$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x + 2 = y + \left(2 y + 6\right)$$
$$x + 2 = 3 y + 6$$
We move the free summand 2 from the left part to the right part performing the sign change
$$x = \left(3 y + 6\right) - 2$$
$$x = 3 y + 4$$
Let's try the obtained element x to 2-th equation
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
We get:
$$4 \left(- 2 y + \left(3 y + 4\right)\right) + \left(- y - 2 \left(3 y + 4\right)\right) = 2 - 2 \left(y + 2 \left(3 y + 4\right)\right)$$
$$8 - 3 y = - 14 y - 14$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$14 y + \left(8 - 3 y\right) = -14$$
$$11 y + 8 = -14$$
We move the free summand 8 from the left part to the right part performing the sign change
$$11 y = -14 - 8$$
$$11 y = -22$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{11 y}{11} = - \frac{22}{11}$$
$$y = -2$$
Because
$$x = 3 y + 4$$
then
$$x = \left(-2\right) 3 + 4$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = -2$$
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
Let's express from equation 1 x
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
Let's move the summand with the variable x from the right part to the left part performing the sign change
$$\left(2 - 5 \left(- 2 x + \frac{y}{5}\right)\right) + \left(6 - 3 \left(3 x + 2\right)\right) = 2 y + 6$$
$$x - y + 2 = 2 y + 6$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x + 2 = y + \left(2 y + 6\right)$$
$$x + 2 = 3 y + 6$$
We move the free summand 2 from the left part to the right part performing the sign change
$$x = \left(3 y + 6\right) - 2$$
$$x = 3 y + 4$$
Let's try the obtained element x to 2-th equation
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
We get:
$$4 \left(- 2 y + \left(3 y + 4\right)\right) + \left(- y - 2 \left(3 y + 4\right)\right) = 2 - 2 \left(y + 2 \left(3 y + 4\right)\right)$$
$$8 - 3 y = - 14 y - 14$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$14 y + \left(8 - 3 y\right) = -14$$
$$11 y + 8 = -14$$
We move the free summand 8 from the left part to the right part performing the sign change
$$11 y = -14 - 8$$
$$11 y = -22$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{11 y}{11} = - \frac{22}{11}$$
$$y = -2$$
Because
$$x = 3 y + 4$$
then
$$x = \left(-2\right) 3 + 4$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = -2$$
Rapid solution
$$x_{1} = -2$$
=
$$-2$$
=
$$y_{1} = -2$$
=
$$-2$$
=
=
$$-2$$
=
-2
$$y_{1} = -2$$
=
$$-2$$
=
-2
Gaussian elimination
Given the system of equations
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
We give the system of equations to the canonical form
$$x - 3 y = 4$$
$$6 x - 7 y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -3 & 4\\6 & -7 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -3 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 6 + 6 & -7 - - 18 & 2 - 4 \cdot 6\end{matrix}\right] = \left[\begin{matrix}0 & 11 & -22\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -3 & 4\\0 & 11 & -22\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\11\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 11 & -22\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{11} & -3 - \frac{\left(-3\right) 11}{11} & 4 - - -6\end{matrix}\right] = \left[\begin{matrix}1 & 0 & -2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & -2\\0 & 11 & -22\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + 2 = 0$$
$$11 x_{2} + 22 = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = -2$$
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
We give the system of equations to the canonical form
$$x - 3 y = 4$$
$$6 x - 7 y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -3 & 4\\6 & -7 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -3 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 6 + 6 & -7 - - 18 & 2 - 4 \cdot 6\end{matrix}\right] = \left[\begin{matrix}0 & 11 & -22\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -3 & 4\\0 & 11 & -22\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\11\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 11 & -22\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{11} & -3 - \frac{\left(-3\right) 11}{11} & 4 - - -6\end{matrix}\right] = \left[\begin{matrix}1 & 0 & -2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & -2\\0 & 11 & -22\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + 2 = 0$$
$$11 x_{2} + 22 = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = -2$$
Cramer's rule
$$2 - 5 \left(- 2 x + \frac{y}{5}\right) = 2 y + 3 \left(3 x + 2\right)$$
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
We give the system of equations to the canonical form
$$x - 3 y = 4$$
$$6 x - 7 y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - 3 x_{2}\\6 x_{1} - 7 x_{2}\end{matrix}\right] = \left[\begin{matrix}4\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -3\\6 & -7\end{matrix}\right] \right)} = 11$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & -3\\2 & -7\end{matrix}\right] \right)}}{11} = -2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 4\\6 & 2\end{matrix}\right] \right)}}{11} = -2$$
$$\left(- 2 x - y\right) + 4 \left(x - 2 y\right) = 2 - 2 \left(2 x + y\right)$$
We give the system of equations to the canonical form
$$x - 3 y = 4$$
$$6 x - 7 y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - 3 x_{2}\\6 x_{1} - 7 x_{2}\end{matrix}\right] = \left[\begin{matrix}4\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -3\\6 & -7\end{matrix}\right] \right)} = 11$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & -3\\2 & -7\end{matrix}\right] \right)}}{11} = -2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 4\\6 & 2\end{matrix}\right] \right)}}{11} = -2$$