Mister Exam
1/6u-1/3v=-3; 0.2u+0.1v=3.9
The solution
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u v - - - = -3 6 3
$$\frac{u}{6} - \frac{v}{3} = -3$$
u v 39 - + -- = -- 5 10 10
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
u/5 + v/10 = 39/10
Detail solution
Given the system of equations
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
Let's express from equation 1 u
$$\frac{u}{6} - \frac{v}{3} = -3$$
Let's move the summand with the variable v from the left part to the right part performing the sign change
$$\frac{u}{6} = \frac{v}{3} - 3$$
$$\frac{u}{6} = \frac{v}{3} - 3$$
Let's divide both parts of the equation by the multiplier of u
$$u = 2 v - 18$$
Let's try the obtained element u to 2-th equation
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
We get:
$$\frac{v}{10} + \frac{2 v - 18}{5} = \frac{39}{10}$$
$$\frac{v}{2} - \frac{18}{5} = \frac{39}{10}$$
We move the free summand -18/5 from the left part to the right part performing the sign change
$$\frac{v}{2} = \frac{18}{5} + \frac{39}{10}$$
$$\frac{v}{2} = \frac{15}{2}$$
Let's divide both parts of the equation by the multiplier of v
$$v = 15$$
Because
$$u = 2 v - 18$$
then
$$u = -18 + 2 \cdot 15$$
$$u = 12$$
The answer:
$$u = 12$$
$$v = 15$$
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
Let's express from equation 1 u
$$\frac{u}{6} - \frac{v}{3} = -3$$
Let's move the summand with the variable v from the left part to the right part performing the sign change
$$\frac{u}{6} = \frac{v}{3} - 3$$
$$\frac{u}{6} = \frac{v}{3} - 3$$
Let's divide both parts of the equation by the multiplier of u
/u\ v |-| -3 + - \6/ 3 --- = ------ 1/6 1/6
$$u = 2 v - 18$$
Let's try the obtained element u to 2-th equation
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
We get:
$$\frac{v}{10} + \frac{2 v - 18}{5} = \frac{39}{10}$$
$$\frac{v}{2} - \frac{18}{5} = \frac{39}{10}$$
We move the free summand -18/5 from the left part to the right part performing the sign change
$$\frac{v}{2} = \frac{18}{5} + \frac{39}{10}$$
$$\frac{v}{2} = \frac{15}{2}$$
Let's divide both parts of the equation by the multiplier of v
/v\ |-| \2/ 15 --- = ----- 1/2 2*1/2
$$v = 15$$
Because
$$u = 2 v - 18$$
then
$$u = -18 + 2 \cdot 15$$
$$u = 12$$
The answer:
$$u = 12$$
$$v = 15$$
Rapid solution
$$u_{1} = 12$$
=
$$12$$
=
$$v_{1} = 15$$
=
$$15$$
=
=
$$12$$
=
12
$$v_{1} = 15$$
=
$$15$$
=
15
Cramer's rule
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
We give the system of equations to the canonical form
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{x_{1}}{6} - \frac{x_{2}}{3}\\\frac{x_{1}}{5} + \frac{x_{2}}{10}\end{matrix}\right] = \left[\begin{matrix}-3\\\frac{39}{10}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3}\\\frac{1}{5} & \frac{1}{10}\end{matrix}\right] \right)} = \frac{1}{12}$$
, 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 )
$$x_{1} = 12 \operatorname{det}{\left(\left[\begin{matrix}-3 & - \frac{1}{3}\\\frac{39}{10} & \frac{1}{10}\end{matrix}\right] \right)} = 12$$
$$x_{2} = 12 \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{6} & -3\\\frac{1}{5} & \frac{39}{10}\end{matrix}\right] \right)} = 15$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
We give the system of equations to the canonical form
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{x_{1}}{6} - \frac{x_{2}}{3}\\\frac{x_{1}}{5} + \frac{x_{2}}{10}\end{matrix}\right] = \left[\begin{matrix}-3\\\frac{39}{10}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3}\\\frac{1}{5} & \frac{1}{10}\end{matrix}\right] \right)} = \frac{1}{12}$$
, 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 )
$$x_{1} = 12 \operatorname{det}{\left(\left[\begin{matrix}-3 & - \frac{1}{3}\\\frac{39}{10} & \frac{1}{10}\end{matrix}\right] \right)} = 12$$
$$x_{2} = 12 \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{6} & -3\\\frac{1}{5} & \frac{39}{10}\end{matrix}\right] \right)} = 15$$
Gaussian elimination
Given the system of equations
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
We give the system of equations to the canonical form
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3} & -3\\\frac{1}{5} & \frac{1}{10} & \frac{39}{10}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{1}{6}\\\frac{1}{5}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3} & -3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{5} - \frac{6}{5 \cdot 6} & \frac{1}{10} - - \frac{2}{5} & \frac{39}{10} - - \frac{18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3} & -3\\0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{3}\\\frac{1}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{6} - \frac{\left(-2\right) 0}{3} & - \frac{1}{3} - \frac{-2}{2 \cdot 3} & -3 - \frac{\left(-2\right) 15}{2 \cdot 3}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{6} & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{6} & 0 & 2\\0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{x_{1}}{6} - 2 = 0$$
$$\frac{x_{2}}{2} - \frac{15}{2} = 0$$
We get the answer:
$$x_{1} = 12$$
$$x_{2} = 15$$
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
We give the system of equations to the canonical form
$$\frac{u}{6} - \frac{v}{3} = -3$$
$$\frac{u}{5} + \frac{v}{10} = \frac{39}{10}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3} & -3\\\frac{1}{5} & \frac{1}{10} & \frac{39}{10}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{1}{6}\\\frac{1}{5}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3} & -3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{5} - \frac{6}{5 \cdot 6} & \frac{1}{10} - - \frac{2}{5} & \frac{39}{10} - - \frac{18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{6} & - \frac{1}{3} & -3\\0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{3}\\\frac{1}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{1}{6} - \frac{\left(-2\right) 0}{3} & - \frac{1}{3} - \frac{-2}{2 \cdot 3} & -3 - \frac{\left(-2\right) 15}{2 \cdot 3}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{6} & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{1}{6} & 0 & 2\\0 & \frac{1}{2} & \frac{15}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{x_{1}}{6} - 2 = 0$$
$$\frac{x_{2}}{2} - \frac{15}{2} = 0$$
We get the answer:
$$x_{1} = 12$$
$$x_{2} = 15$$