Mister Exam
2x+4y=30; X-5y=8
The solution
Detail solution
Given the system of equations
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
Let's express from equation 1 x
$$2 x + 4 y = 30$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 30 - 4 y$$
$$2 x = 30 - 4 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{30 - 4 y}{2}$$
$$x = 15 - 2 y$$
Let's try the obtained element x to 2-th equation
$$x - 5 y = 8$$
We get:
$$- 5 y + \left(15 - 2 y\right) = 8$$
$$15 - 7 y = 8$$
We move the free summand 15 from the left part to the right part performing the sign change
$$- 7 y = -15 + 8$$
$$- 7 y = -7$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 7 y}{-7} = - \frac{7}{-7}$$
$$y = 1$$
Because
$$x = 15 - 2 y$$
then
$$x = -2 + 15$$
$$x = 13$$
The answer:
$$x = 13$$
$$y = 1$$
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
Let's express from equation 1 x
$$2 x + 4 y = 30$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 30 - 4 y$$
$$2 x = 30 - 4 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{30 - 4 y}{2}$$
$$x = 15 - 2 y$$
Let's try the obtained element x to 2-th equation
$$x - 5 y = 8$$
We get:
$$- 5 y + \left(15 - 2 y\right) = 8$$
$$15 - 7 y = 8$$
We move the free summand 15 from the left part to the right part performing the sign change
$$- 7 y = -15 + 8$$
$$- 7 y = -7$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 7 y}{-7} = - \frac{7}{-7}$$
$$y = 1$$
Because
$$x = 15 - 2 y$$
then
$$x = -2 + 15$$
$$x = 13$$
The answer:
$$x = 13$$
$$y = 1$$
Rapid solution
$$x_{1} = 13$$
=
$$13$$
=
$$y_{1} = 1$$
=
$$1$$
=
=
$$13$$
=
13
$$y_{1} = 1$$
=
$$1$$
=
1
Gaussian elimination
Given the system of equations
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
We give the system of equations to the canonical form
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 4 & 30\\1 & -5 & 8\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 4 & 30\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{2}{2} & -5 - \frac{4}{2} & 8 - \frac{30}{2}\end{matrix}\right] = \left[\begin{matrix}0 & -7 & -7\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 4 & 30\\0 & -7 & -7\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\-7\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -7 & -7\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{7} & 4 - - -4 & 30 - - -4\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 26\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 26\\0 & -7 & -7\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 26 = 0$$
$$7 - 7 x_{2} = 0$$
We get the answer:
$$x_{1} = 13$$
$$x_{2} = 1$$
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
We give the system of equations to the canonical form
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 4 & 30\\1 & -5 & 8\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 4 & 30\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{2}{2} & -5 - \frac{4}{2} & 8 - \frac{30}{2}\end{matrix}\right] = \left[\begin{matrix}0 & -7 & -7\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 4 & 30\\0 & -7 & -7\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\-7\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -7 & -7\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{7} & 4 - - -4 & 30 - - -4\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 26\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 26\\0 & -7 & -7\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 26 = 0$$
$$7 - 7 x_{2} = 0$$
We get the answer:
$$x_{1} = 13$$
$$x_{2} = 1$$
Cramer's rule
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
We give the system of equations to the canonical form
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 4 x_{2}\\x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}30\\8\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 4\\1 & -5\end{matrix}\right] \right)} = -14$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}30 & 4\\8 & -5\end{matrix}\right] \right)}}{14} = 13$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 30\\1 & 8\end{matrix}\right] \right)}}{14} = 1$$
$$x - 5 y = 8$$
We give the system of equations to the canonical form
$$2 x + 4 y = 30$$
$$x - 5 y = 8$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 4 x_{2}\\x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}30\\8\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 4\\1 & -5\end{matrix}\right] \right)} = -14$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}30 & 4\\8 & -5\end{matrix}\right] \right)}}{14} = 13$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 30\\1 & 8\end{matrix}\right] \right)}}{14} = 1$$