Mister Exam
5m+n=-16; m+5n=-4
The solution
Detail solution
Given the system of equations
$$5 m + n = -16$$
$$m + 5 n = -4$$
Let's express from equation 1 m
$$5 m + n = -16$$
Let's move the summand with the variable n from the left part to the right part performing the sign change
$$5 m = - n - 16$$
$$5 m = - n - 16$$
Let's divide both parts of the equation by the multiplier of m
$$\frac{5 m}{5} = \frac{- n - 16}{5}$$
$$m = - \frac{n}{5} - \frac{16}{5}$$
Let's try the obtained element m to 2-th equation
$$m + 5 n = -4$$
We get:
$$5 n + \left(- \frac{n}{5} - \frac{16}{5}\right) = -4$$
$$\frac{24 n}{5} - \frac{16}{5} = -4$$
We move the free summand -16/5 from the left part to the right part performing the sign change
$$\frac{24 n}{5} = -4 + \frac{16}{5}$$
$$\frac{24 n}{5} = - \frac{4}{5}$$
Let's divide both parts of the equation by the multiplier of n
$$\frac{\frac{24}{5} n}{\frac{24}{5}} = - \frac{4}{\frac{24}{5} \cdot 5}$$
$$n = - \frac{1}{6}$$
Because
$$m = - \frac{n}{5} - \frac{16}{5}$$
then
$$m = - \frac{16}{5} - - \frac{1}{30}$$
$$m = - \frac{19}{6}$$
The answer:
$$m = - \frac{19}{6}$$
$$n = - \frac{1}{6}$$
$$5 m + n = -16$$
$$m + 5 n = -4$$
Let's express from equation 1 m
$$5 m + n = -16$$
Let's move the summand with the variable n from the left part to the right part performing the sign change
$$5 m = - n - 16$$
$$5 m = - n - 16$$
Let's divide both parts of the equation by the multiplier of m
$$\frac{5 m}{5} = \frac{- n - 16}{5}$$
$$m = - \frac{n}{5} - \frac{16}{5}$$
Let's try the obtained element m to 2-th equation
$$m + 5 n = -4$$
We get:
$$5 n + \left(- \frac{n}{5} - \frac{16}{5}\right) = -4$$
$$\frac{24 n}{5} - \frac{16}{5} = -4$$
We move the free summand -16/5 from the left part to the right part performing the sign change
$$\frac{24 n}{5} = -4 + \frac{16}{5}$$
$$\frac{24 n}{5} = - \frac{4}{5}$$
Let's divide both parts of the equation by the multiplier of n
$$\frac{\frac{24}{5} n}{\frac{24}{5}} = - \frac{4}{\frac{24}{5} \cdot 5}$$
$$n = - \frac{1}{6}$$
Because
$$m = - \frac{n}{5} - \frac{16}{5}$$
then
$$m = - \frac{16}{5} - - \frac{1}{30}$$
$$m = - \frac{19}{6}$$
The answer:
$$m = - \frac{19}{6}$$
$$n = - \frac{1}{6}$$
Rapid solution
$$m_{1} = - \frac{19}{6}$$
=
$$- \frac{19}{6}$$
=
$$n_{1} = - \frac{1}{6}$$
=
$$- \frac{1}{6}$$
=
=
$$- \frac{19}{6}$$
=
-3.16666666666667
$$n_{1} = - \frac{1}{6}$$
=
$$- \frac{1}{6}$$
=
-0.166666666666667
Cramer's rule
$$5 m + n = -16$$
$$m + 5 n = -4$$
We give the system of equations to the canonical form
$$5 m + n = -16$$
$$m + 5 n = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + x_{2}\\x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}-16\\-4\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 1\\1 & 5\end{matrix}\right] \right)} = 24$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-16 & 1\\-4 & 5\end{matrix}\right] \right)}}{24} = - \frac{19}{6}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -16\\1 & -4\end{matrix}\right] \right)}}{24} = - \frac{1}{6}$$
$$m + 5 n = -4$$
We give the system of equations to the canonical form
$$5 m + n = -16$$
$$m + 5 n = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + x_{2}\\x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}-16\\-4\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 1\\1 & 5\end{matrix}\right] \right)} = 24$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-16 & 1\\-4 & 5\end{matrix}\right] \right)}}{24} = - \frac{19}{6}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -16\\1 & -4\end{matrix}\right] \right)}}{24} = - \frac{1}{6}$$
Gaussian elimination
Given the system of equations
$$5 m + n = -16$$
$$m + 5 n = -4$$
We give the system of equations to the canonical form
$$5 m + n = -16$$
$$m + 5 n = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 1 & -16\\1 & 5 & -4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 1 & -16\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{5}{5} & \frac{-1}{5} + 5 & -4 - - \frac{16}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & -16\\0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\\frac{24}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 5}{24} & 1 - \frac{5 \cdot 24}{5 \cdot 24} & -16 - - \frac{1}{6}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & - \frac{95}{6}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & - \frac{95}{6}\\0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} + \frac{95}{6} = 0$$
$$\frac{24 x_{2}}{5} + \frac{4}{5} = 0$$
We get the answer:
$$x_{1} = - \frac{19}{6}$$
$$x_{2} = - \frac{1}{6}$$
$$5 m + n = -16$$
$$m + 5 n = -4$$
We give the system of equations to the canonical form
$$5 m + n = -16$$
$$m + 5 n = -4$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 1 & -16\\1 & 5 & -4\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 1 & -16\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{5}{5} & \frac{-1}{5} + 5 & -4 - - \frac{16}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & -16\\0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\\frac{24}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{0 \cdot 5}{24} & 1 - \frac{5 \cdot 24}{5 \cdot 24} & -16 - - \frac{1}{6}\end{matrix}\right] = \left[\begin{matrix}5 & 0 & - \frac{95}{6}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & - \frac{95}{6}\\0 & \frac{24}{5} & - \frac{4}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} + \frac{95}{6} = 0$$
$$\frac{24 x_{2}}{5} + \frac{4}{5} = 0$$
We get the answer:
$$x_{1} = - \frac{19}{6}$$
$$x_{2} = - \frac{1}{6}$$
Numerical answer
[src]
m1 = -3.166666666666667 n1 = -0.1666666666666667
m1 = -3.166666666666667 n1 = -0.1666666666666667