Mister Exam
Х+3у=-14; 2х-5у=38
The solution
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x + 3*y = -14
$$x + 3 y = -14$$
2*x - 5*y = 38
$$2 x - 5 y = 38$$
2*x - 5*y = 38
Detail solution
Given the system of equations
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
Let's express from equation 1 x
$$x + 3 y = -14$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = - 3 y - 14$$
$$x = - 3 y - 14$$
Let's try the obtained element x to 2-th equation
$$2 x - 5 y = 38$$
We get:
$$- 5 y + 2 \left(- 3 y - 14\right) = 38$$
$$- 11 y - 28 = 38$$
We move the free summand -28 from the left part to the right part performing the sign change
$$- 11 y = 28 + 38$$
$$- 11 y = 66$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 11 y}{-11} = \frac{66}{-11}$$
$$y = -6$$
Because
$$x = - 3 y - 14$$
then
$$x = -14 - -18$$
$$x = 4$$
The answer:
$$x = 4$$
$$y = -6$$
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
Let's express from equation 1 x
$$x + 3 y = -14$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = - 3 y - 14$$
$$x = - 3 y - 14$$
Let's try the obtained element x to 2-th equation
$$2 x - 5 y = 38$$
We get:
$$- 5 y + 2 \left(- 3 y - 14\right) = 38$$
$$- 11 y - 28 = 38$$
We move the free summand -28 from the left part to the right part performing the sign change
$$- 11 y = 28 + 38$$
$$- 11 y = 66$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 11 y}{-11} = \frac{66}{-11}$$
$$y = -6$$
Because
$$x = - 3 y - 14$$
then
$$x = -14 - -18$$
$$x = 4$$
The answer:
$$x = 4$$
$$y = -6$$
Rapid solution
$$x_{1} = 4$$
=
$$4$$
=
$$y_{1} = -6$$
=
$$-6$$
=
=
$$4$$
=
4
$$y_{1} = -6$$
=
$$-6$$
=
-6
Cramer's rule
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
We give the system of equations to the canonical form
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 3 x_{2}\\2 x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}-14\\38\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 3\\2 & -5\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-14 & 3\\38 & -5\end{matrix}\right] \right)}}{11} = 4$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -14\\2 & 38\end{matrix}\right] \right)}}{11} = -6$$
$$2 x - 5 y = 38$$
We give the system of equations to the canonical form
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 3 x_{2}\\2 x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}-14\\38\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 3\\2 & -5\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-14 & 3\\38 & -5\end{matrix}\right] \right)}}{11} = 4$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -14\\2 & 38\end{matrix}\right] \right)}}{11} = -6$$
Gaussian elimination
Given the system of equations
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
We give the system of equations to the canonical form
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 3 & -14\\2 & -5 & 38\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 3 & -14\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & - 2 \cdot 3 - 5 & 38 - - 28\end{matrix}\right] = \left[\begin{matrix}0 & -11 & 66\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 3 & -14\\0 & -11 & 66\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\-11\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -11 & 66\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{11} & 3 - - -3 & -14 - \frac{\left(-3\right) 66}{11}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 4\\0 & -11 & 66\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 4 = 0$$
$$- 11 x_{2} - 66 = 0$$
We get the answer:
$$x_{1} = 4$$
$$x_{2} = -6$$
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
We give the system of equations to the canonical form
$$x + 3 y = -14$$
$$2 x - 5 y = 38$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 3 & -14\\2 & -5 & 38\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 3 & -14\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & - 2 \cdot 3 - 5 & 38 - - 28\end{matrix}\right] = \left[\begin{matrix}0 & -11 & 66\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 3 & -14\\0 & -11 & 66\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}3\\-11\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -11 & 66\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-3\right) 0}{11} & 3 - - -3 & -14 - \frac{\left(-3\right) 66}{11}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 4\\0 & -11 & 66\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 4 = 0$$
$$- 11 x_{2} - 66 = 0$$
We get the answer:
$$x_{1} = 4$$
$$x_{2} = -6$$