Mister Exam
0.4(x+y)=12; 0.6(x-y)=9
The solution
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2*(x + y)
--------- = 12
5
$$\frac{2 \left(x + y\right)}{5} = 12$$
3*(x - y)
--------- = 9
5
$$\frac{3 \left(x - y\right)}{5} = 9$$
3*(x - y)/5 = 9
Detail solution
Given the system of equations
$$\frac{2 \left(x + y\right)}{5} = 12$$
$$\frac{3 \left(x - y\right)}{5} = 9$$
Let's express from equation 1 x
$$\frac{2 \left(x + y\right)}{5} = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{2 x}{5} = \left(\frac{2 x}{5} - \frac{2 \left(x + y\right)}{5}\right) + 12$$
$$\frac{2 x}{5} = 12 - \frac{2 y}{5}$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{2}{5} x}{\frac{2}{5}} = \frac{12 - \frac{2 y}{5}}{\frac{2}{5}}$$
$$x = 30 - y$$
Let's try the obtained element x to 2-th equation
$$\frac{3 \left(x - y\right)}{5} = 9$$
We get:
$$\frac{3 \left(- y + \left(30 - y\right)\right)}{5} = 9$$
$$18 - \frac{6 y}{5} = 9$$
We move the free summand 18 from the left part to the right part performing the sign change
$$- \frac{6 y}{5} = -18 + 9$$
$$- \frac{6 y}{5} = -9$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{6}{5} y}{- \frac{6}{5}} = - \frac{9}{- \frac{6}{5}}$$
$$y = \frac{15}{2}$$
Because
$$x = 30 - y$$
then
$$x = 30 - \frac{15}{2}$$
$$x = \frac{45}{2}$$
The answer:
$$x = \frac{45}{2}$$
$$y = \frac{15}{2}$$
$$\frac{2 \left(x + y\right)}{5} = 12$$
$$\frac{3 \left(x - y\right)}{5} = 9$$
Let's express from equation 1 x
$$\frac{2 \left(x + y\right)}{5} = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{2 x}{5} = \left(\frac{2 x}{5} - \frac{2 \left(x + y\right)}{5}\right) + 12$$
$$\frac{2 x}{5} = 12 - \frac{2 y}{5}$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{2}{5} x}{\frac{2}{5}} = \frac{12 - \frac{2 y}{5}}{\frac{2}{5}}$$
$$x = 30 - y$$
Let's try the obtained element x to 2-th equation
$$\frac{3 \left(x - y\right)}{5} = 9$$
We get:
$$\frac{3 \left(- y + \left(30 - y\right)\right)}{5} = 9$$
$$18 - \frac{6 y}{5} = 9$$
We move the free summand 18 from the left part to the right part performing the sign change
$$- \frac{6 y}{5} = -18 + 9$$
$$- \frac{6 y}{5} = -9$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{6}{5} y}{- \frac{6}{5}} = - \frac{9}{- \frac{6}{5}}$$
$$y = \frac{15}{2}$$
Because
$$x = 30 - y$$
then
$$x = 30 - \frac{15}{2}$$
$$x = \frac{45}{2}$$
The answer:
$$x = \frac{45}{2}$$
$$y = \frac{15}{2}$$
Rapid solution
$$x_{1} = \frac{45}{2}$$
=
$$\frac{45}{2}$$
=
$$y_{1} = \frac{15}{2}$$
=
$$\frac{15}{2}$$
=
=
$$\frac{45}{2}$$
=
22.5
$$y_{1} = \frac{15}{2}$$
=
$$\frac{15}{2}$$
=
7.5
Gaussian elimination
Given the system of equations
$$\frac{2 \left(x + y\right)}{5} = 12$$
$$\frac{3 \left(x - y\right)}{5} = 9$$
We give the system of equations to the canonical form
$$\frac{2 x}{5} + \frac{2 y}{5} = 12$$
$$\frac{3 x}{5} - \frac{3 y}{5} = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{2}{5} & \frac{2}{5} & 12\\\frac{3}{5} & - \frac{3}{5} & 9\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{2}{5}\\\frac{3}{5}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{2}{5} & \frac{2}{5} & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{3}{5} - \frac{2 \cdot 3}{2 \cdot 5} & - \frac{2 \cdot 3}{2 \cdot 5} - \frac{3}{5} & 9 - \frac{3 \cdot 12}{2}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{6}{5} & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{2}{5} & \frac{2}{5} & 12\\0 & - \frac{6}{5} & -9\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}\frac{2}{5}\\- \frac{6}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{6}{5} & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{2}{5} - \frac{\left(-1\right) 0}{3} & \frac{2}{5} - - \frac{-2}{5} & 12 - - -3\end{matrix}\right] = \left[\begin{matrix}\frac{2}{5} & 0 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{2}{5} & 0 & 9\\0 & - \frac{6}{5} & -9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{2 x_{1}}{5} - 9 = 0$$
$$9 - \frac{6 x_{2}}{5} = 0$$
We get the answer:
$$x_{1} = \frac{45}{2}$$
$$x_{2} = \frac{15}{2}$$
$$\frac{2 \left(x + y\right)}{5} = 12$$
$$\frac{3 \left(x - y\right)}{5} = 9$$
We give the system of equations to the canonical form
$$\frac{2 x}{5} + \frac{2 y}{5} = 12$$
$$\frac{3 x}{5} - \frac{3 y}{5} = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{2}{5} & \frac{2}{5} & 12\\\frac{3}{5} & - \frac{3}{5} & 9\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{2}{5}\\\frac{3}{5}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{2}{5} & \frac{2}{5} & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{3}{5} - \frac{2 \cdot 3}{2 \cdot 5} & - \frac{2 \cdot 3}{2 \cdot 5} - \frac{3}{5} & 9 - \frac{3 \cdot 12}{2}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{6}{5} & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{2}{5} & \frac{2}{5} & 12\\0 & - \frac{6}{5} & -9\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}\frac{2}{5}\\- \frac{6}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{6}{5} & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{2}{5} - \frac{\left(-1\right) 0}{3} & \frac{2}{5} - - \frac{-2}{5} & 12 - - -3\end{matrix}\right] = \left[\begin{matrix}\frac{2}{5} & 0 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{2}{5} & 0 & 9\\0 & - \frac{6}{5} & -9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{2 x_{1}}{5} - 9 = 0$$
$$9 - \frac{6 x_{2}}{5} = 0$$
We get the answer:
$$x_{1} = \frac{45}{2}$$
$$x_{2} = \frac{15}{2}$$
Cramer's rule
$$\frac{2 \left(x + y\right)}{5} = 12$$
$$\frac{3 \left(x - y\right)}{5} = 9$$
We give the system of equations to the canonical form
$$\frac{2 x}{5} + \frac{2 y}{5} = 12$$
$$\frac{3 x}{5} - \frac{3 y}{5} = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{2 x_{1}}{5} + \frac{2 x_{2}}{5}\\\frac{3 x_{1}}{5} - \frac{3 x_{2}}{5}\end{matrix}\right] = \left[\begin{matrix}12\\9\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{2}{5} & \frac{2}{5}\\\frac{3}{5} & - \frac{3}{5}\end{matrix}\right] \right)} = - \frac{12}{25}$$
, 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 )
$$x_{1} = - \frac{25 \operatorname{det}{\left(\left[\begin{matrix}12 & \frac{2}{5}\\9 & - \frac{3}{5}\end{matrix}\right] \right)}}{12} = \frac{45}{2}$$
$$x_{2} = - \frac{25 \operatorname{det}{\left(\left[\begin{matrix}\frac{2}{5} & 12\\\frac{3}{5} & 9\end{matrix}\right] \right)}}{12} = \frac{15}{2}$$
$$\frac{3 \left(x - y\right)}{5} = 9$$
We give the system of equations to the canonical form
$$\frac{2 x}{5} + \frac{2 y}{5} = 12$$
$$\frac{3 x}{5} - \frac{3 y}{5} = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{2 x_{1}}{5} + \frac{2 x_{2}}{5}\\\frac{3 x_{1}}{5} - \frac{3 x_{2}}{5}\end{matrix}\right] = \left[\begin{matrix}12\\9\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{2}{5} & \frac{2}{5}\\\frac{3}{5} & - \frac{3}{5}\end{matrix}\right] \right)} = - \frac{12}{25}$$
, 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 )
$$x_{1} = - \frac{25 \operatorname{det}{\left(\left[\begin{matrix}12 & \frac{2}{5}\\9 & - \frac{3}{5}\end{matrix}\right] \right)}}{12} = \frac{45}{2}$$
$$x_{2} = - \frac{25 \operatorname{det}{\left(\left[\begin{matrix}\frac{2}{5} & 12\\\frac{3}{5} & 9\end{matrix}\right] \right)}}{12} = \frac{15}{2}$$