Mister Exam
9*x-11*y=5; 6*y-12*x=-8
The solution
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9*x - 11*y = 5
$$9 x - 11 y = 5$$
6*y - 12*x = -8
$$- 12 x + 6 y = -8$$
-12*x + 6*y = -8
Detail solution
Given the system of equations
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
Let's express from equation 1 x
$$9 x - 11 y = 5$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$9 x = 11 y + 5$$
$$9 x = 11 y + 5$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{9 x}{9} = \frac{11 y + 5}{9}$$
$$x = \frac{11 y}{9} + \frac{5}{9}$$
Let's try the obtained element x to 2-th equation
$$- 12 x + 6 y = -8$$
We get:
$$6 y - 12 \left(\frac{11 y}{9} + \frac{5}{9}\right) = -8$$
$$- \frac{26 y}{3} - \frac{20}{3} = -8$$
We move the free summand -20/3 from the left part to the right part performing the sign change
$$- \frac{26 y}{3} = -8 + \frac{20}{3}$$
$$- \frac{26 y}{3} = - \frac{4}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{26}{3} y}{- \frac{26}{3}} = - \frac{4}{\left(- \frac{26}{3}\right) 3}$$
$$y = \frac{2}{13}$$
Because
$$x = \frac{11 y}{9} + \frac{5}{9}$$
then
$$x = \frac{2 \cdot 11}{9 \cdot 13} + \frac{5}{9}$$
$$x = \frac{29}{39}$$
The answer:
$$x = \frac{29}{39}$$
$$y = \frac{2}{13}$$
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
Let's express from equation 1 x
$$9 x - 11 y = 5$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$9 x = 11 y + 5$$
$$9 x = 11 y + 5$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{9 x}{9} = \frac{11 y + 5}{9}$$
$$x = \frac{11 y}{9} + \frac{5}{9}$$
Let's try the obtained element x to 2-th equation
$$- 12 x + 6 y = -8$$
We get:
$$6 y - 12 \left(\frac{11 y}{9} + \frac{5}{9}\right) = -8$$
$$- \frac{26 y}{3} - \frac{20}{3} = -8$$
We move the free summand -20/3 from the left part to the right part performing the sign change
$$- \frac{26 y}{3} = -8 + \frac{20}{3}$$
$$- \frac{26 y}{3} = - \frac{4}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{26}{3} y}{- \frac{26}{3}} = - \frac{4}{\left(- \frac{26}{3}\right) 3}$$
$$y = \frac{2}{13}$$
Because
$$x = \frac{11 y}{9} + \frac{5}{9}$$
then
$$x = \frac{2 \cdot 11}{9 \cdot 13} + \frac{5}{9}$$
$$x = \frac{29}{39}$$
The answer:
$$x = \frac{29}{39}$$
$$y = \frac{2}{13}$$
Rapid solution
$$x_{1} = \frac{29}{39}$$
=
$$\frac{29}{39}$$
=
$$y_{1} = \frac{2}{13}$$
=
$$\frac{2}{13}$$
=
=
$$\frac{29}{39}$$
=
0.743589743589744
$$y_{1} = \frac{2}{13}$$
=
$$\frac{2}{13}$$
=
0.153846153846154
Cramer's rule
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
We give the system of equations to the canonical form
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 x_{1} - 11 x_{2}\\- 12 x_{1} + 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}5\\-8\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}9 & -11\\-12 & 6\end{matrix}\right] \right)} = -78$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -11\\-8 & 6\end{matrix}\right] \right)}}{78} = \frac{29}{39}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & 5\\-12 & -8\end{matrix}\right] \right)}}{78} = \frac{2}{13}$$
$$- 12 x + 6 y = -8$$
We give the system of equations to the canonical form
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 x_{1} - 11 x_{2}\\- 12 x_{1} + 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}5\\-8\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}9 & -11\\-12 & 6\end{matrix}\right] \right)} = -78$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & -11\\-8 & 6\end{matrix}\right] \right)}}{78} = \frac{29}{39}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & 5\\-12 & -8\end{matrix}\right] \right)}}{78} = \frac{2}{13}$$
Gaussian elimination
Given the system of equations
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
We give the system of equations to the canonical form
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 & -11 & 5\\-12 & 6 & -8\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}9\\-12\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}9 & -11 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-12 - \frac{\left(-4\right) 9}{3} & 6 - - \frac{-44}{3} & -8 - \frac{\left(-4\right) 5}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & -11 & 5\\0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-11\\- \frac{26}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}9 - \frac{0 \cdot 33}{26} & -11 - - 11 & 5 - - \frac{22}{13}\end{matrix}\right] = \left[\begin{matrix}9 & 0 & \frac{87}{13}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & 0 & \frac{87}{13}\\0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$9 x_{1} - \frac{87}{13} = 0$$
$$\frac{4}{3} - \frac{26 x_{2}}{3} = 0$$
We get the answer:
$$x_{1} = \frac{29}{39}$$
$$x_{2} = \frac{2}{13}$$
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
We give the system of equations to the canonical form
$$9 x - 11 y = 5$$
$$- 12 x + 6 y = -8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 & -11 & 5\\-12 & 6 & -8\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}9\\-12\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}9 & -11 & 5\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-12 - \frac{\left(-4\right) 9}{3} & 6 - - \frac{-44}{3} & -8 - \frac{\left(-4\right) 5}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & -11 & 5\\0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-11\\- \frac{26}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}9 - \frac{0 \cdot 33}{26} & -11 - - 11 & 5 - - \frac{22}{13}\end{matrix}\right] = \left[\begin{matrix}9 & 0 & \frac{87}{13}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & 0 & \frac{87}{13}\\0 & - \frac{26}{3} & - \frac{4}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$9 x_{1} - \frac{87}{13} = 0$$
$$\frac{4}{3} - \frac{26 x_{2}}{3} = 0$$
We get the answer:
$$x_{1} = \frac{29}{39}$$
$$x_{2} = \frac{2}{13}$$
Numerical answer
[src]
x1 = 0.7435897435897436 y1 = 0.1538461538461538
x1 = 0.7435897435897436 y1 = 0.1538461538461538