Mister Exam
5x−3y=13; 2x+y=3
The solution
Detail solution
Given the system of equations
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
Let's express from equation 1 x
$$5 x - 3 y = 13$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 3 y + 13$$
$$5 x = 3 y + 13$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{3 y + 13}{5}$$
$$x = \frac{3 y}{5} + \frac{13}{5}$$
Let's try the obtained element x to 2-th equation
$$2 x + y = 3$$
We get:
$$y + 2 \left(\frac{3 y}{5} + \frac{13}{5}\right) = 3$$
$$\frac{11 y}{5} + \frac{26}{5} = 3$$
We move the free summand 26/5 from the left part to the right part performing the sign change
$$\frac{11 y}{5} = - \frac{26}{5} + 3$$
$$\frac{11 y}{5} = - \frac{11}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{11}{5} y}{\frac{11}{5}} = - \frac{11}{\frac{11}{5} \cdot 5}$$
$$y = -1$$
Because
$$x = \frac{3 y}{5} + \frac{13}{5}$$
then
$$x = \frac{\left(-1\right) 3}{5} + \frac{13}{5}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = -1$$
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
Let's express from equation 1 x
$$5 x - 3 y = 13$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$5 x = 3 y + 13$$
$$5 x = 3 y + 13$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{5 x}{5} = \frac{3 y + 13}{5}$$
$$x = \frac{3 y}{5} + \frac{13}{5}$$
Let's try the obtained element x to 2-th equation
$$2 x + y = 3$$
We get:
$$y + 2 \left(\frac{3 y}{5} + \frac{13}{5}\right) = 3$$
$$\frac{11 y}{5} + \frac{26}{5} = 3$$
We move the free summand 26/5 from the left part to the right part performing the sign change
$$\frac{11 y}{5} = - \frac{26}{5} + 3$$
$$\frac{11 y}{5} = - \frac{11}{5}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{11}{5} y}{\frac{11}{5}} = - \frac{11}{\frac{11}{5} \cdot 5}$$
$$y = -1$$
Because
$$x = \frac{3 y}{5} + \frac{13}{5}$$
then
$$x = \frac{\left(-1\right) 3}{5} + \frac{13}{5}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = -1$$
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = -1$$
=
$$-1$$
=
=
$$2$$
=
2
$$y_{1} = -1$$
=
$$-1$$
=
-1
Gaussian elimination
Given the system of equations
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
We give the system of equations to the canonical form
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -3 & 13\\2 & 1 & 3\end{matrix}\right]$$
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
$$\left[\begin{matrix}5 & -3 & 13\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & 1 - - \frac{6}{5} & 3 - \frac{2 \cdot 13}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -3 & 13\\0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{11}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{11} & -3 - \frac{\left(-15\right) 11}{5 \cdot 11} & 13 - - -3\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 10\\0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 10 = 0$$
$$\frac{11 x_{2}}{5} + \frac{11}{5} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = -1$$
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
We give the system of equations to the canonical form
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 & -3 & 13\\2 & 1 & 3\end{matrix}\right]$$
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
$$\left[\begin{matrix}5 & -3 & 13\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{2 \cdot 5}{5} & 1 - - \frac{6}{5} & 3 - \frac{2 \cdot 13}{5}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & -3 & 13\\0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-3\\\frac{11}{5}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{\left(-15\right) 0}{11} & -3 - \frac{\left(-15\right) 11}{5 \cdot 11} & 13 - - -3\end{matrix}\right] = \left[\begin{matrix}5 & 0 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}5 & 0 & 10\\0 & \frac{11}{5} & - \frac{11}{5}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$5 x_{1} - 10 = 0$$
$$\frac{11 x_{2}}{5} + \frac{11}{5} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = -1$$
Cramer's rule
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
We give the system of equations to the canonical form
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 3 x_{2}\\2 x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}13\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -3\\2 & 1\end{matrix}\right] \right)} = 11$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}13 & -3\\3 & 1\end{matrix}\right] \right)}}{11} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 13\\2 & 3\end{matrix}\right] \right)}}{11} = -1$$
$$2 x + y = 3$$
We give the system of equations to the canonical form
$$5 x - 3 y = 13$$
$$2 x + y = 3$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}5 x_{1} - 3 x_{2}\\2 x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}13\\3\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}5 & -3\\2 & 1\end{matrix}\right] \right)} = 11$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}13 & -3\\3 & 1\end{matrix}\right] \right)}}{11} = 2$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}5 & 13\\2 & 3\end{matrix}\right] \right)}}{11} = -1$$