Mister Exam
m+1/5-3n-5/10=-2; m-3/6+5n-9/4=-2,5
The solution
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m + 1/5 - 3*n - 1/2 = -2
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
m - 1/2 + 5*n - 9/4 = -5/2
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
5*n + m - 1/2 - 9/4 = -5/2
Detail solution
Given the system of equations
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
Let's express from equation 1 m
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
Let's move the summand with the variable n from the left part to the right part performing the sign change
$$m - \frac{3}{10} = 3 n - 2$$
$$m - \frac{3}{10} = 3 n - 2$$
We move the free summand -3/10 from the left part to the right part performing the sign change
$$m = \left(3 n - 2\right) + \frac{3}{10}$$
$$m = 3 n - \frac{17}{10}$$
Let's try the obtained element m to 2-th equation
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
We get:
$$\left(5 n + \left(\left(3 n - \frac{17}{10}\right) - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
$$8 n - \frac{89}{20} = - \frac{5}{2}$$
We move the free summand -89/20 from the left part to the right part performing the sign change
$$8 n = - \frac{5}{2} + \frac{89}{20}$$
$$8 n = \frac{39}{20}$$
Let's divide both parts of the equation by the multiplier of n
$$\frac{8 n}{8} = \frac{39}{8 \cdot 20}$$
$$n = \frac{39}{160}$$
Because
$$m = 3 n - \frac{17}{10}$$
then
$$m = - \frac{17}{10} + \frac{3 \cdot 39}{160}$$
$$m = - \frac{31}{32}$$
The answer:
$$m = - \frac{31}{32}$$
$$n = \frac{39}{160}$$
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
Let's express from equation 1 m
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
Let's move the summand with the variable n from the left part to the right part performing the sign change
$$m - \frac{3}{10} = 3 n - 2$$
$$m - \frac{3}{10} = 3 n - 2$$
We move the free summand -3/10 from the left part to the right part performing the sign change
$$m = \left(3 n - 2\right) + \frac{3}{10}$$
$$m = 3 n - \frac{17}{10}$$
Let's try the obtained element m to 2-th equation
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
We get:
$$\left(5 n + \left(\left(3 n - \frac{17}{10}\right) - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
$$8 n - \frac{89}{20} = - \frac{5}{2}$$
We move the free summand -89/20 from the left part to the right part performing the sign change
$$8 n = - \frac{5}{2} + \frac{89}{20}$$
$$8 n = \frac{39}{20}$$
Let's divide both parts of the equation by the multiplier of n
$$\frac{8 n}{8} = \frac{39}{8 \cdot 20}$$
$$n = \frac{39}{160}$$
Because
$$m = 3 n - \frac{17}{10}$$
then
$$m = - \frac{17}{10} + \frac{3 \cdot 39}{160}$$
$$m = - \frac{31}{32}$$
The answer:
$$m = - \frac{31}{32}$$
$$n = \frac{39}{160}$$
Rapid solution
$$m_{1} = - \frac{31}{32}$$
=
$$- \frac{31}{32}$$
=
$$n_{1} = \frac{39}{160}$$
=
$$\frac{39}{160}$$
=
=
$$- \frac{31}{32}$$
=
-0.96875
$$n_{1} = \frac{39}{160}$$
=
$$\frac{39}{160}$$
=
0.24375
Cramer's rule
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
We give the system of equations to the canonical form
$$m - 3 n = - \frac{17}{10}$$
$$m + 5 n = \frac{1}{4}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - 3 x_{2}\\x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}- \frac{17}{10}\\\frac{1}{4}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -3\\1 & 5\end{matrix}\right] \right)} = 8$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}- \frac{17}{10} & -3\\\frac{1}{4} & 5\end{matrix}\right] \right)}}{8} = - \frac{31}{32}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & - \frac{17}{10}\\1 & \frac{1}{4}\end{matrix}\right] \right)}}{8} = \frac{39}{160}$$
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
We give the system of equations to the canonical form
$$m - 3 n = - \frac{17}{10}$$
$$m + 5 n = \frac{1}{4}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - 3 x_{2}\\x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}- \frac{17}{10}\\\frac{1}{4}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & -3\\1 & 5\end{matrix}\right] \right)} = 8$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}- \frac{17}{10} & -3\\\frac{1}{4} & 5\end{matrix}\right] \right)}}{8} = - \frac{31}{32}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & - \frac{17}{10}\\1 & \frac{1}{4}\end{matrix}\right] \right)}}{8} = \frac{39}{160}$$
Gaussian elimination
Given the system of equations
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
We give the system of equations to the canonical form
$$m - 3 n = - \frac{17}{10}$$
$$m + 5 n = \frac{1}{4}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -3 & - \frac{17}{10}\\1 & 5 & \frac{1}{4}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -3 & - \frac{17}{10}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & 5 - -3 & \frac{1}{4} - - \frac{17}{10}\end{matrix}\right] = \left[\begin{matrix}0 & 8 & \frac{39}{20}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -3 & - \frac{17}{10}\\0 & 8 & \frac{39}{20}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\8\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 8 & \frac{39}{20}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{8} & -3 - \frac{\left(-3\right) 8}{8} & - \frac{17}{10} - \frac{\left(-3\right) 39}{8 \cdot 20}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & - \frac{31}{32}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & - \frac{31}{32}\\0 & 8 & \frac{39}{20}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + \frac{31}{32} = 0$$
$$8 x_{2} - \frac{39}{20} = 0$$
We get the answer:
$$x_{1} = - \frac{31}{32}$$
$$x_{2} = \frac{39}{160}$$
$$\left(- 3 n + \left(m + \frac{1}{5}\right)\right) - \frac{1}{2} = -2$$
$$\left(5 n + \left(m - \frac{1}{2}\right)\right) - \frac{9}{4} = - \frac{5}{2}$$
We give the system of equations to the canonical form
$$m - 3 n = - \frac{17}{10}$$
$$m + 5 n = \frac{1}{4}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -3 & - \frac{17}{10}\\1 & 5 & \frac{1}{4}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -3 & - \frac{17}{10}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & 5 - -3 & \frac{1}{4} - - \frac{17}{10}\end{matrix}\right] = \left[\begin{matrix}0 & 8 & \frac{39}{20}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -3 & - \frac{17}{10}\\0 & 8 & \frac{39}{20}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\8\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 8 & \frac{39}{20}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{8} & -3 - \frac{\left(-3\right) 8}{8} & - \frac{17}{10} - \frac{\left(-3\right) 39}{8 \cdot 20}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & - \frac{31}{32}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & - \frac{31}{32}\\0 & 8 & \frac{39}{20}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + \frac{31}{32} = 0$$
$$8 x_{2} - \frac{39}{20} = 0$$
We get the answer:
$$x_{1} = - \frac{31}{32}$$
$$x_{2} = \frac{39}{160}$$