Mister Exam
5х-2y=6; 7x+2y=6
The solution
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5*x - 2*y = 6
$$5 x - 2 y = 6$$
7*x + 2*y = 6
$$7 x + 2 y = 6$$
7*x + 2*y = 6
Detail solution
Given the system of equations
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
Let's express from equation 1 x
$$5 x - 2 y = 6$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 2 y + 6$$
$$5 x = 2 y + 6$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{2 y + 6}{5}$$
$$x = \frac{2 y}{5} + \frac{6}{5}$$
Let's try the obtained element x to 2-th equation
$$7 x + 2 y = 6$$
We get:
$$2 y + 7 \left(\frac{2 y}{5} + \frac{6}{5}\right) = 6$$
$$\frac{24 y}{5} + \frac{42}{5} = 6$$
We move the free summand 42/5 from the left part to the right part performing the sign change
$$\frac{24 y}{5} = - \frac{42}{5} + 6$$
$$\frac{24 y}{5} = - \frac{12}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{24}{5} y}{\frac{24}{5}} = - \frac{12}{\frac{24}{5} \cdot 5}$$
$$y = - \frac{1}{2}$$
Because
$$x = \frac{2 y}{5} + \frac{6}{5}$$
then
$$x = \frac{\left(-1\right) 2}{2 \cdot 5} + \frac{6}{5}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = - \frac{1}{2}$$
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
Let's express from equation 1 x
$$5 x - 2 y = 6$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 2 y + 6$$
$$5 x = 2 y + 6$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{2 y + 6}{5}$$
$$x = \frac{2 y}{5} + \frac{6}{5}$$
Let's try the obtained element x to 2-th equation
$$7 x + 2 y = 6$$
We get:
$$2 y + 7 \left(\frac{2 y}{5} + \frac{6}{5}\right) = 6$$
$$\frac{24 y}{5} + \frac{42}{5} = 6$$
We move the free summand 42/5 from the left part to the right part performing the sign change
$$\frac{24 y}{5} = - \frac{42}{5} + 6$$
$$\frac{24 y}{5} = - \frac{12}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{24}{5} y}{\frac{24}{5}} = - \frac{12}{\frac{24}{5} \cdot 5}$$
$$y = - \frac{1}{2}$$
Because
$$x = \frac{2 y}{5} + \frac{6}{5}$$
then
$$x = \frac{\left(-1\right) 2}{2 \cdot 5} + \frac{6}{5}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = - \frac{1}{2}$$
Rapid solution
$$x_{1} = 1$$
=
$$1$$
=
$$y_{1} = - \frac{1}{2}$$
=
$$- \frac{1}{2}$$
=
=
$$1$$
=
1
$$y_{1} = - \frac{1}{2}$$
=
$$- \frac{1}{2}$$
=
-0.5
Gaussian elimination
Given the system of equations
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
We give the system of equations to the canonical form
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -2 & 6\\7 & 2 & 6\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -2 & 6\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{5 \cdot 7}{5} & 2 - - \frac{14}{5} & 6 - \frac{6 \cdot 7}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -2 & 6\\0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{24}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{12} & -2 - \frac{\left(-5\right) 24}{5 \cdot 12} & 6 - - -1\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 5\\0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 5 = 0$$
$$\frac{24 x_{2}}{5} + \frac{12}{5} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = - \frac{1}{2}$$
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
We give the system of equations to the canonical form
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -2 & 6\\7 & 2 & 6\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -2 & 6\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{5 \cdot 7}{5} & 2 - - \frac{14}{5} & 6 - \frac{6 \cdot 7}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -2 & 6\\0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\\frac{24}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{12} & -2 - \frac{\left(-5\right) 24}{5 \cdot 12} & 6 - - -1\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 5\\0 & \frac{24}{5} & - \frac{12}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 5 = 0$$
$$\frac{24 x_{2}}{5} + \frac{12}{5} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = - \frac{1}{2}$$
Cramer's rule
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
We give the system of equations to the canonical form
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 2 x_{2}\\7 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}6\\6\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -2\\7 & 2\end{matrix}\right] \right)} = 24$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}6 & -2\\6 & 2\end{matrix}\right] \right)}}{24} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 6\\7 & 6\end{matrix}\right] \right)}}{24} = - \frac{1}{2}$$
$$7 x + 2 y = 6$$
We give the system of equations to the canonical form
$$5 x - 2 y = 6$$
$$7 x + 2 y = 6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 2 x_{2}\\7 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}6\\6\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -2\\7 & 2\end{matrix}\right] \right)} = 24$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}6 & -2\\6 & 2\end{matrix}\right] \right)}}{24} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 6\\7 & 6\end{matrix}\right] \right)}}{24} = - \frac{1}{2}$$