Mister Exam
2х-у=0; 3х+2у=14
The solution
Detail solution
Given the system of equations
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
Let's express from equation 1 x
$$2 x - y = 0$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = y$$
$$2 x = y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{y}{2}$$
$$x = \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$3 x + 2 y = 14$$
We get:
$$3 \frac{y}{2} + 2 y = 14$$
$$\frac{7 y}{2} = 14$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{7}{2} y}{\frac{7}{2}} = \frac{14}{\frac{7}{2}}$$
$$y = 4$$
Because
$$x = \frac{y}{2}$$
then
$$x = \frac{4}{2}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = 4$$
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
Let's express from equation 1 x
$$2 x - y = 0$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = y$$
$$2 x = y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{y}{2}$$
$$x = \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$3 x + 2 y = 14$$
We get:
$$3 \frac{y}{2} + 2 y = 14$$
$$\frac{7 y}{2} = 14$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{7}{2} y}{\frac{7}{2}} = \frac{14}{\frac{7}{2}}$$
$$y = 4$$
Because
$$x = \frac{y}{2}$$
then
$$x = \frac{4}{2}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = 4$$
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = 4$$
=
$$4$$
=
=
$$2$$
=
2
$$y_{1} = 4$$
=
$$4$$
=
4
Gaussian elimination
Given the system of equations
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
We give the system of equations to the canonical form
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -1 & 0\\3 & 2 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -1 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 2 - - \frac{3}{2} & 14 - \frac{0 \cdot 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{7}{2} & 14\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -1 & 0\\0 & \frac{7}{2} & 14\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\\frac{7}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{7}{2} & 14\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-2\right) 0}{7} & -1 - \frac{\left(-2\right) 7}{2 \cdot 7} & - \frac{\left(-2\right) 14}{7}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 4\\0 & \frac{7}{2} & 14\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 4 = 0$$
$$\frac{7 x_{2}}{2} - 14 = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 4$$
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
We give the system of equations to the canonical form
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -1 & 0\\3 & 2 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -1 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 2 - - \frac{3}{2} & 14 - \frac{0 \cdot 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{7}{2} & 14\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -1 & 0\\0 & \frac{7}{2} & 14\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\\frac{7}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{7}{2} & 14\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-2\right) 0}{7} & -1 - \frac{\left(-2\right) 7}{2 \cdot 7} & - \frac{\left(-2\right) 14}{7}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 4\\0 & \frac{7}{2} & 14\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 4 = 0$$
$$\frac{7 x_{2}}{2} - 14 = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 4$$
Cramer's rule
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
We give the system of equations to the canonical form
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - x_{2}\\3 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}0\\14\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -1\\3 & 2\end{matrix}\right] \right)} = 7$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}0 & -1\\14 & 2\end{matrix}\right] \right)}}{7} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 0\\3 & 14\end{matrix}\right] \right)}}{7} = 4$$
$$3 x + 2 y = 14$$
We give the system of equations to the canonical form
$$2 x - y = 0$$
$$3 x + 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - x_{2}\\3 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}0\\14\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -1\\3 & 2\end{matrix}\right] \right)} = 7$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}0 & -1\\14 & 2\end{matrix}\right] \right)}}{7} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 0\\3 & 14\end{matrix}\right] \right)}}{7} = 4$$