Mister Exam
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-
How to use it?
- 5х-7у=9; 6х+5у=-16
- x2+7xy=-6; 9y2-xy=10
-
х^2-xy=-8; y^2-xy=24
- log7(36−y^2)=log7(36−a^2x^2); x^2+y^2=2x+6y
-
Identical expressions
- 5х-7у= nine ; 6х+5у=- sixteen
- 5х minus 7у equally 9; 6х plus 5у equally minus 16
- 5х minus 7у equally nine ; 6х plus 5у equally minus sixteen
-
Similar expressions
- 5х-7у=9; 6х-5у=-16
- 5х-7у=9; 6х+5у=+16
- 5х+7у=9; 6х+5у=-16
5х-7у=9; 6х+5у=-16
The solution
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5*x - 7*y = 9
$$5 x - 7 y = 9$$
6*x + 5*y = -16
$$6 x + 5 y = -16$$
6*x + 5*y = -16
Detail solution
Given the system of equations
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Let's express from equation 1 x
$$5 x - 7 y = 9$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 7 y + 9$$
$$5 x = 7 y + 9$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{7 y + 9}{5}$$
$$x = \frac{7 y}{5} + \frac{9}{5}$$
Let's try the obtained element x to 2-th equation
$$6 x + 5 y = -16$$
We get:
$$5 y + 6 \left(\frac{7 y}{5} + \frac{9}{5}\right) = -16$$
$$\frac{67 y}{5} + \frac{54}{5} = -16$$
We move the free summand 54/5 from the left part to the right part performing the sign change
$$\frac{67 y}{5} = -16 - \frac{54}{5}$$
$$\frac{67 y}{5} = - \frac{134}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{67}{5} y}{\frac{67}{5}} = - \frac{134}{5 \frac{67}{5}}$$
$$y = -2$$
Because
$$x = \frac{7 y}{5} + \frac{9}{5}$$
then
$$x = \frac{\left(-2\right) 7}{5} + \frac{9}{5}$$
$$x = -1$$
The answer:
$$x = -1$$
$$y = -2$$
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Let's express from equation 1 x
$$5 x - 7 y = 9$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 7 y + 9$$
$$5 x = 7 y + 9$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{7 y + 9}{5}$$
$$x = \frac{7 y}{5} + \frac{9}{5}$$
Let's try the obtained element x to 2-th equation
$$6 x + 5 y = -16$$
We get:
$$5 y + 6 \left(\frac{7 y}{5} + \frac{9}{5}\right) = -16$$
$$\frac{67 y}{5} + \frac{54}{5} = -16$$
We move the free summand 54/5 from the left part to the right part performing the sign change
$$\frac{67 y}{5} = -16 - \frac{54}{5}$$
$$\frac{67 y}{5} = - \frac{134}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{67}{5} y}{\frac{67}{5}} = - \frac{134}{5 \frac{67}{5}}$$
$$y = -2$$
Because
$$x = \frac{7 y}{5} + \frac{9}{5}$$
then
$$x = \frac{\left(-2\right) 7}{5} + \frac{9}{5}$$
$$x = -1$$
The answer:
$$x = -1$$
$$y = -2$$
Rapid solution
$$x_{1} = -1$$
=
$$-1$$
=
$$y_{1} = -2$$
=
$$-2$$
=
=
$$-1$$
=
-1
$$y_{1} = -2$$
=
$$-2$$
=
-2
Gaussian elimination
Given the system of equations
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
We give the system of equations to the canonical form
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -7 & 9\\6 & 5 & -16\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -7 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{5 \cdot 6}{5} & 5 - - \frac{42}{5} & -16 - \frac{6 \cdot 9}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -7 & 9\\0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\\frac{67}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-35\right) 0}{67} & -7 - \frac{\left(-35\right) 67}{5 \cdot 67} & 9 - - -14\end{matrix}\right] = \left[\begin{matrix}5 & 0 & -5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5\\0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} + 5 = 0$$
$$\frac{67 x_{2}}{5} + \frac{134}{5} = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = -2$$
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
We give the system of equations to the canonical form
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -7 & 9\\6 & 5 & -16\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -7 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{5 \cdot 6}{5} & 5 - - \frac{42}{5} & -16 - \frac{6 \cdot 9}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -7 & 9\\0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-7\\\frac{67}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-35\right) 0}{67} & -7 - \frac{\left(-35\right) 67}{5 \cdot 67} & 9 - - -14\end{matrix}\right] = \left[\begin{matrix}5 & 0 & -5\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & -5\\0 & \frac{67}{5} & - \frac{134}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} + 5 = 0$$
$$\frac{67 x_{2}}{5} + \frac{134}{5} = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = -2$$
Cramer's rule
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
We give the system of equations to the canonical form
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 7 x_{2}\\6 x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}9\\-16\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -7\\6 & 5\end{matrix}\right] \right)} = 67$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & -7\\-16 & 5\end{matrix}\right] \right)}}{67} = -1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 9\\6 & -16\end{matrix}\right] \right)}}{67} = -2$$
$$6 x + 5 y = -16$$
We give the system of equations to the canonical form
$$5 x - 7 y = 9$$
$$6 x + 5 y = -16$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 7 x_{2}\\6 x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}9\\-16\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -7\\6 & 5\end{matrix}\right] \right)} = 67$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & -7\\-16 & 5\end{matrix}\right] \right)}}{67} = -1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 9\\6 & -16\end{matrix}\right] \right)}}{67} = -2$$