Mister Exam
x/3-(y-2x)/5=1/3; y/2+5/6=(x+y)/3
The solution
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x y - 2*x - - ------- = 1/3 3 5
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
y 5 x + y - + - = ----- 2 6 3
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
y/2 + 5/6 = (x + y)/3
Detail solution
Given the system of equations
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
Let's express from equation 1 x
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{x}{3} + \frac{2 x}{5} = \left(\frac{2 x}{5} + \frac{- 2 x + y}{5}\right) + \frac{1}{3}$$
$$\frac{11 x}{15} = \frac{y}{5} + \frac{1}{3}$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{11}{15} x}{\frac{11}{15}} = \frac{\frac{y}{5} + \frac{1}{3}}{\frac{11}{15}}$$
$$x = \frac{3 y}{11} + \frac{5}{11}$$
Let's try the obtained element x to 2-th equation
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
We get:
$$\frac{y}{2} + \frac{5}{6} = \frac{y + \left(\frac{3 y}{11} + \frac{5}{11}\right)}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{14 y}{33} + \frac{5}{33}$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$- \frac{14 y}{33} + \left(\frac{y}{2} + \frac{5}{6}\right) = \frac{5}{33}$$
$$\frac{5 y}{66} + \frac{5}{6} = \frac{5}{33}$$
We move the free summand 5/6 from the left part to the right part performing the sign change
$$\frac{5 y}{66} = - \frac{5}{6} + \frac{5}{33}$$
$$\frac{5 y}{66} = - \frac{15}{22}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{5}{66} y}{\frac{5}{66}} = - \frac{15}{\frac{5}{66} \cdot 22}$$
$$y = -9$$
Because
$$x = \frac{3 y}{11} + \frac{5}{11}$$
then
$$x = \frac{\left(-9\right) 3}{11} + \frac{5}{11}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = -9$$
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
Let's express from equation 1 x
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{x}{3} + \frac{2 x}{5} = \left(\frac{2 x}{5} + \frac{- 2 x + y}{5}\right) + \frac{1}{3}$$
$$\frac{11 x}{15} = \frac{y}{5} + \frac{1}{3}$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{11}{15} x}{\frac{11}{15}} = \frac{\frac{y}{5} + \frac{1}{3}}{\frac{11}{15}}$$
$$x = \frac{3 y}{11} + \frac{5}{11}$$
Let's try the obtained element x to 2-th equation
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
We get:
$$\frac{y}{2} + \frac{5}{6} = \frac{y + \left(\frac{3 y}{11} + \frac{5}{11}\right)}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{14 y}{33} + \frac{5}{33}$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$- \frac{14 y}{33} + \left(\frac{y}{2} + \frac{5}{6}\right) = \frac{5}{33}$$
$$\frac{5 y}{66} + \frac{5}{6} = \frac{5}{33}$$
We move the free summand 5/6 from the left part to the right part performing the sign change
$$\frac{5 y}{66} = - \frac{5}{6} + \frac{5}{33}$$
$$\frac{5 y}{66} = - \frac{15}{22}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{5}{66} y}{\frac{5}{66}} = - \frac{15}{\frac{5}{66} \cdot 22}$$
$$y = -9$$
Because
$$x = \frac{3 y}{11} + \frac{5}{11}$$
then
$$x = \frac{\left(-9\right) 3}{11} + \frac{5}{11}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = -9$$
Rapid solution
$$x_{1} = -2$$
=
$$-2$$
=
$$y_{1} = -9$$
=
$$-9$$
=
=
$$-2$$
=
-2
$$y_{1} = -9$$
=
$$-9$$
=
-9
Cramer's rule
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
We give the system of equations to the canonical form
$$\frac{11 x}{15} - \frac{y}{5} = \frac{1}{3}$$
$$- \frac{x}{3} + \frac{y}{6} = - \frac{5}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{11 x_{1}}{15} - \frac{x_{2}}{5}\\- \frac{x_{1}}{3} + \frac{x_{2}}{6}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{3}\\- \frac{5}{6}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5}\\- \frac{1}{3} & \frac{1}{6}\end{matrix}\right] \right)} = \frac{1}{18}$$
, 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 )
$$x_{1} = 18 \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{3} & - \frac{1}{5}\\- \frac{5}{6} & \frac{1}{6}\end{matrix}\right] \right)} = -2$$
$$x_{2} = 18 \operatorname{det}{\left(\left[\begin{matrix}\frac{11}{15} & \frac{1}{3}\\- \frac{1}{3} & - \frac{5}{6}\end{matrix}\right] \right)} = -9$$
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
We give the system of equations to the canonical form
$$\frac{11 x}{15} - \frac{y}{5} = \frac{1}{3}$$
$$- \frac{x}{3} + \frac{y}{6} = - \frac{5}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{11 x_{1}}{15} - \frac{x_{2}}{5}\\- \frac{x_{1}}{3} + \frac{x_{2}}{6}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{3}\\- \frac{5}{6}\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5}\\- \frac{1}{3} & \frac{1}{6}\end{matrix}\right] \right)} = \frac{1}{18}$$
, 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 )
$$x_{1} = 18 \operatorname{det}{\left(\left[\begin{matrix}\frac{1}{3} & - \frac{1}{5}\\- \frac{5}{6} & \frac{1}{6}\end{matrix}\right] \right)} = -2$$
$$x_{2} = 18 \operatorname{det}{\left(\left[\begin{matrix}\frac{11}{15} & \frac{1}{3}\\- \frac{1}{3} & - \frac{5}{6}\end{matrix}\right] \right)} = -9$$
Gaussian elimination
Given the system of equations
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
We give the system of equations to the canonical form
$$\frac{11 x}{15} - \frac{y}{5} = \frac{1}{3}$$
$$- \frac{x}{3} + \frac{y}{6} = - \frac{5}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5} & \frac{1}{3}\\- \frac{1}{3} & \frac{1}{6} & - \frac{5}{6}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{11}{15}\\- \frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5} & \frac{1}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{1}{3} - \frac{\left(-5\right) 11}{11 \cdot 15} & \frac{1}{6} - - \frac{-1}{11} & - \frac{5}{6} - \frac{-5}{3 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5} & \frac{1}{3}\\0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{5}\\\frac{5}{66}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{11}{15} - \frac{\left(-66\right) 0}{25} & - \frac{1}{5} - \frac{\left(-66\right) 5}{25 \cdot 66} & \frac{1}{3} - - \frac{-9}{5}\end{matrix}\right] = \left[\begin{matrix}\frac{11}{15} & 0 & - \frac{22}{15}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{11}{15} & 0 & - \frac{22}{15}\\0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{11 x_{1}}{15} + \frac{22}{15} = 0$$
$$\frac{5 x_{2}}{66} + \frac{15}{22} = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = -9$$
$$\frac{x}{3} - \frac{- 2 x + y}{5} = \frac{1}{3}$$
$$\frac{y}{2} + \frac{5}{6} = \frac{x + y}{3}$$
We give the system of equations to the canonical form
$$\frac{11 x}{15} - \frac{y}{5} = \frac{1}{3}$$
$$- \frac{x}{3} + \frac{y}{6} = - \frac{5}{6}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5} & \frac{1}{3}\\- \frac{1}{3} & \frac{1}{6} & - \frac{5}{6}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{11}{15}\\- \frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5} & \frac{1}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{1}{3} - \frac{\left(-5\right) 11}{11 \cdot 15} & \frac{1}{6} - - \frac{-1}{11} & - \frac{5}{6} - \frac{-5}{3 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{11}{15} & - \frac{1}{5} & \frac{1}{3}\\0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{5}\\\frac{5}{66}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{11}{15} - \frac{\left(-66\right) 0}{25} & - \frac{1}{5} - \frac{\left(-66\right) 5}{25 \cdot 66} & \frac{1}{3} - - \frac{-9}{5}\end{matrix}\right] = \left[\begin{matrix}\frac{11}{15} & 0 & - \frac{22}{15}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{11}{15} & 0 & - \frac{22}{15}\\0 & \frac{5}{66} & - \frac{15}{22}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{11 x_{1}}{15} + \frac{22}{15} = 0$$
$$\frac{5 x_{2}}{66} + \frac{15}{22} = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = -9$$