Mister Exam
3х-5у=-18; 2х+5у=13
The solution
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3*x - 5*y = -18
$$3 x - 5 y = -18$$
2*x + 5*y = 13
$$2 x + 5 y = 13$$
2*x + 5*y = 13
Detail solution
Given the system of equations
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
Let's express from equation 1 x
$$3 x - 5 y = -18$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 5 y - 18$$
$$3 x = 5 y - 18$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{5 y - 18}{3}$$
$$x = \frac{5 y}{3} - 6$$
Let's try the obtained element x to 2-th equation
$$2 x + 5 y = 13$$
We get:
$$5 y + 2 \left(\frac{5 y}{3} - 6\right) = 13$$
$$\frac{25 y}{3} - 12 = 13$$
We move the free summand -12 from the left part to the right part performing the sign change
$$\frac{25 y}{3} = 12 + 13$$
$$\frac{25 y}{3} = 25$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{25}{3} y}{\frac{25}{3}} = \frac{25}{\frac{25}{3}}$$
$$y = 3$$
Because
$$x = \frac{5 y}{3} - 6$$
then
$$x = -6 + \frac{3 \cdot 5}{3}$$
$$x = -1$$
The answer:
$$x = -1$$
$$y = 3$$
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
Let's express from equation 1 x
$$3 x - 5 y = -18$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 5 y - 18$$
$$3 x = 5 y - 18$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{5 y - 18}{3}$$
$$x = \frac{5 y}{3} - 6$$
Let's try the obtained element x to 2-th equation
$$2 x + 5 y = 13$$
We get:
$$5 y + 2 \left(\frac{5 y}{3} - 6\right) = 13$$
$$\frac{25 y}{3} - 12 = 13$$
We move the free summand -12 from the left part to the right part performing the sign change
$$\frac{25 y}{3} = 12 + 13$$
$$\frac{25 y}{3} = 25$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{25}{3} y}{\frac{25}{3}} = \frac{25}{\frac{25}{3}}$$
$$y = 3$$
Because
$$x = \frac{5 y}{3} - 6$$
then
$$x = -6 + \frac{3 \cdot 5}{3}$$
$$x = -1$$
The answer:
$$x = -1$$
$$y = 3$$
Rapid solution
$$x_{1} = -1$$
=
$$-1$$
=
$$y_{1} = 3$$
=
$$3$$
=
=
$$-1$$
=
-1
$$y_{1} = 3$$
=
$$3$$
=
3
Gaussian elimination
Given the system of equations
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
We give the system of equations to the canonical form
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -5 & -18\\2 & 5 & 13\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & -18\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & 5 - - \frac{10}{3} & 13 - - 12\end{matrix}\right] = \left[\begin{matrix}0 & \frac{25}{3} & 25\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -5 & -18\\0 & \frac{25}{3} & 25\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{25}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{25}{3} & 25\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{5} & -5 - \frac{\left(-3\right) 25}{3 \cdot 5} & -18 - \frac{\left(-3\right) 25}{5}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -3\\0 & \frac{25}{3} & 25\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} + 3 = 0$$
$$\frac{25 x_{2}}{3} - 25 = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = 3$$
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
We give the system of equations to the canonical form
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -5 & -18\\2 & 5 & 13\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & -18\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 3}{3} & 5 - - \frac{10}{3} & 13 - - 12\end{matrix}\right] = \left[\begin{matrix}0 & \frac{25}{3} & 25\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -5 & -18\\0 & \frac{25}{3} & 25\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\\frac{25}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{25}{3} & 25\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{5} & -5 - \frac{\left(-3\right) 25}{3 \cdot 5} & -18 - \frac{\left(-3\right) 25}{5}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -3\\0 & \frac{25}{3} & 25\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} + 3 = 0$$
$$\frac{25 x_{2}}{3} - 25 = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = 3$$
Cramer's rule
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
We give the system of equations to the canonical form
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 5 x_{2}\\2 x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}-18\\13\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -5\\2 & 5\end{matrix}\right] \right)} = 25$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-18 & -5\\13 & 5\end{matrix}\right] \right)}}{25} = -1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -18\\2 & 13\end{matrix}\right] \right)}}{25} = 3$$
$$2 x + 5 y = 13$$
We give the system of equations to the canonical form
$$3 x - 5 y = -18$$
$$2 x + 5 y = 13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 5 x_{2}\\2 x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}-18\\13\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -5\\2 & 5\end{matrix}\right] \right)} = 25$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-18 & -5\\13 & 5\end{matrix}\right] \right)}}{25} = -1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -18\\2 & 13\end{matrix}\right] \right)}}{25} = 3$$