Mister Exam
5x2+y=12; 9x2-y=2
The solution
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5*x2 + y = 12
$$5 x_{2} + y = 12$$
9*x2 - y = 2
$$9 x_{2} - y = 2$$
9*x2 - y = 2
Detail solution
Given the system of equations
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
Let's express from equation 1 x2
$$5 x_{2} + y = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x_{2} = 12 - y$$
$$5 x_{2} = 12 - y$$
Let's divide both parts of the equation by the multiplier of x2
$$\frac{5 x_{2}}{5} = \frac{12 - y}{5}$$
$$x_{2} = \frac{12}{5} - \frac{y}{5}$$
Let's try the obtained element x2 to 2-th equation
$$9 x_{2} - y = 2$$
We get:
$$- y + 9 \left(\frac{12}{5} - \frac{y}{5}\right) = 2$$
$$\frac{108}{5} - \frac{14 y}{5} = 2$$
We move the free summand 108/5 from the left part to the right part performing the sign change
$$- \frac{14 y}{5} = - \frac{108}{5} + 2$$
$$- \frac{14 y}{5} = - \frac{98}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{14}{5} y}{- \frac{14}{5}} = - \frac{98}{\left(- \frac{14}{5}\right) 5}$$
$$y = 7$$
Because
$$x_{2} = \frac{12}{5} - \frac{y}{5}$$
then
$$x_{2} = \frac{12}{5} - \frac{7}{5}$$
$$x_{2} = 1$$
The answer:
$$x_{2} = 1$$
$$y = 7$$
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
Let's express from equation 1 x2
$$5 x_{2} + y = 12$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x_{2} = 12 - y$$
$$5 x_{2} = 12 - y$$
Let's divide both parts of the equation by the multiplier of x2
$$\frac{5 x_{2}}{5} = \frac{12 - y}{5}$$
$$x_{2} = \frac{12}{5} - \frac{y}{5}$$
Let's try the obtained element x2 to 2-th equation
$$9 x_{2} - y = 2$$
We get:
$$- y + 9 \left(\frac{12}{5} - \frac{y}{5}\right) = 2$$
$$\frac{108}{5} - \frac{14 y}{5} = 2$$
We move the free summand 108/5 from the left part to the right part performing the sign change
$$- \frac{14 y}{5} = - \frac{108}{5} + 2$$
$$- \frac{14 y}{5} = - \frac{98}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{14}{5} y}{- \frac{14}{5}} = - \frac{98}{\left(- \frac{14}{5}\right) 5}$$
$$y = 7$$
Because
$$x_{2} = \frac{12}{5} - \frac{y}{5}$$
then
$$x_{2} = \frac{12}{5} - \frac{7}{5}$$
$$x_{2} = 1$$
The answer:
$$x_{2} = 1$$
$$y = 7$$
Rapid solution
$$x_{21} = 1$$
=
$$1$$
=
$$y_{1} = 7$$
=
$$7$$
=
=
$$1$$
=
1
$$y_{1} = 7$$
=
$$7$$
=
7
Cramer's rule
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
We give the system of equations to the canonical form
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + x_{2}\\9 x_{1} - x_{2}\end{matrix}\right] = \left[\begin{matrix}12\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 1\\9 & -1\end{matrix}\right] \right)} = -14$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}12 & 1\\2 & -1\end{matrix}\right] \right)}}{14} = 1$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 12\\9 & 2\end{matrix}\right] \right)}}{14} = 7$$
$$9 x_{2} - y = 2$$
We give the system of equations to the canonical form
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} + x_{2}\\9 x_{1} - x_{2}\end{matrix}\right] = \left[\begin{matrix}12\\2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & 1\\9 & -1\end{matrix}\right] \right)} = -14$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}12 & 1\\2 & -1\end{matrix}\right] \right)}}{14} = 1$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 12\\9 & 2\end{matrix}\right] \right)}}{14} = 7$$
Gaussian elimination
Given the system of equations
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
We give the system of equations to the canonical form
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 1 & 12\\9 & -1 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\9\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 1 & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}9 - \frac{5 \cdot 9}{5} & \frac{\left(-1\right) 9}{5} - 1 & 2 - \frac{9 \cdot 12}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & 12\\0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{14}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{14} & 1 - - -1 & 12 - - -7\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 5\\0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 5 = 0$$
$$\frac{98}{5} - \frac{14 x_{2}}{5} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = 7$$
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
We give the system of equations to the canonical form
$$5 x_{2} + y = 12$$
$$9 x_{2} - y = 2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & 1 & 12\\9 & -1 & 2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\9\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & 1 & 12\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}9 - \frac{5 \cdot 9}{5} & \frac{\left(-1\right) 9}{5} - 1 & 2 - \frac{9 \cdot 12}{5}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 1 & 12\\0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{14}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{14} & 1 - - -1 & 12 - - -7\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 5\\0 & - \frac{14}{5} & - \frac{98}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 5 = 0$$
$$\frac{98}{5} - \frac{14 x_{2}}{5} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = 7$$