Mister Exam
5х-3у=30; -2х+3у=-21
The solution
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5*x - 3*y = 30
$$5 x - 3 y = 30$$
-2*x + 3*y = -21
$$- 2 x + 3 y = -21$$
-2*x + 3*y = -21
Detail solution
Given the system of equations
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
Let's express from equation 1 x
$$5 x - 3 y = 30$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 3 y + 30$$
$$5 x = 3 y + 30$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{3 y + 30}{5}$$
$$x = \frac{3 y}{5} + 6$$
Let's try the obtained element x to 2-th equation
$$- 2 x + 3 y = -21$$
We get:
$$3 y - 2 \left(\frac{3 y}{5} + 6\right) = -21$$
$$\frac{9 y}{5} - 12 = -21$$
We move the free summand -12 from the left part to the right part performing the sign change
$$\frac{9 y}{5} = -21 + 12$$
$$\frac{9 y}{5} = -9$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{9}{5} y}{\frac{9}{5}} = - \frac{9}{\frac{9}{5}}$$
$$y = -5$$
Because
$$x = \frac{3 y}{5} + 6$$
then
$$x = \frac{\left(-5\right) 3}{5} + 6$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = -5$$
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
Let's express from equation 1 x
$$5 x - 3 y = 30$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 3 y + 30$$
$$5 x = 3 y + 30$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{3 y + 30}{5}$$
$$x = \frac{3 y}{5} + 6$$
Let's try the obtained element x to 2-th equation
$$- 2 x + 3 y = -21$$
We get:
$$3 y - 2 \left(\frac{3 y}{5} + 6\right) = -21$$
$$\frac{9 y}{5} - 12 = -21$$
We move the free summand -12 from the left part to the right part performing the sign change
$$\frac{9 y}{5} = -21 + 12$$
$$\frac{9 y}{5} = -9$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{9}{5} y}{\frac{9}{5}} = - \frac{9}{\frac{9}{5}}$$
$$y = -5$$
Because
$$x = \frac{3 y}{5} + 6$$
then
$$x = \frac{\left(-5\right) 3}{5} + 6$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = -5$$
Rapid solution
$$x_{1} = 3$$
=
$$3$$
=
$$y_{1} = -5$$
=
$$-5$$
=
=
$$3$$
=
3
$$y_{1} = -5$$
=
$$-5$$
=
-5
Gaussian elimination
Given the system of equations
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
We give the system of equations to the canonical form
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -3 & 30\\-2 & 3 & -21\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\-2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -3 & 30\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-2 - \frac{\left(-2\right) 5}{5} & 3 - - \frac{-6}{5} & -21 - \frac{\left(-2\right) 30}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{9}{5} & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -3 & 30\\0 & \frac{9}{5} & -9\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{9}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{9}{5} & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{3} & -3 - \frac{\left(-5\right) 9}{3 \cdot 5} & 30 - - -15\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 15\\0 & \frac{9}{5} & -9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{9 x_{2}}{5} + 9 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = -5$$
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
We give the system of equations to the canonical form
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -3 & 30\\-2 & 3 & -21\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}5\\-2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}5 & -3 & 30\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-2 - \frac{\left(-2\right) 5}{5} & 3 - - \frac{-6}{5} & -21 - \frac{\left(-2\right) 30}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{9}{5} & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -3 & 30\\0 & \frac{9}{5} & -9\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{9}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{9}{5} & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-5\right) 0}{3} & -3 - \frac{\left(-5\right) 9}{3 \cdot 5} & 30 - - -15\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 15\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 15\\0 & \frac{9}{5} & -9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 15 = 0$$
$$\frac{9 x_{2}}{5} + 9 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = -5$$
Cramer's rule
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
We give the system of equations to the canonical form
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 3 x_{2}\\- 2 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}30\\-21\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -3\\-2 & 3\end{matrix}\right] \right)} = 9$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}30 & -3\\-21 & 3\end{matrix}\right] \right)}}{9} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 30\\-2 & -21\end{matrix}\right] \right)}}{9} = -5$$
$$- 2 x + 3 y = -21$$
We give the system of equations to the canonical form
$$5 x - 3 y = 30$$
$$- 2 x + 3 y = -21$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 3 x_{2}\\- 2 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}30\\-21\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -3\\-2 & 3\end{matrix}\right] \right)} = 9$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}30 & -3\\-21 & 3\end{matrix}\right] \right)}}{9} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 30\\-2 & -21\end{matrix}\right] \right)}}{9} = -5$$