Mister Exam
Х+2у=7; 3х+4у=11
The solution
Detail solution
Given the system of equations
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
Let's express from equation 1 x
$$x + 2 y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = 7 - 2 y$$
$$x = 7 - 2 y$$
Let's try the obtained element x to 2-th equation
$$3 x + 4 y = 11$$
We get:
$$4 y + 3 \left(7 - 2 y\right) = 11$$
$$21 - 2 y = 11$$
We move the free summand 21 from the left part to the right part performing the sign change
$$- 2 y = -21 + 11$$
$$- 2 y = -10$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 2 y}{-2} = - \frac{10}{-2}$$
$$y = 5$$
Because
$$x = 7 - 2 y$$
then
$$x = 7 - 10$$
$$x = -3$$
The answer:
$$x = -3$$
$$y = 5$$
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
Let's express from equation 1 x
$$x + 2 y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = 7 - 2 y$$
$$x = 7 - 2 y$$
Let's try the obtained element x to 2-th equation
$$3 x + 4 y = 11$$
We get:
$$4 y + 3 \left(7 - 2 y\right) = 11$$
$$21 - 2 y = 11$$
We move the free summand 21 from the left part to the right part performing the sign change
$$- 2 y = -21 + 11$$
$$- 2 y = -10$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 2 y}{-2} = - \frac{10}{-2}$$
$$y = 5$$
Because
$$x = 7 - 2 y$$
then
$$x = 7 - 10$$
$$x = -3$$
The answer:
$$x = -3$$
$$y = 5$$
Rapid solution
$$x_{1} = -3$$
=
$$-3$$
=
$$y_{1} = 5$$
=
$$5$$
=
=
$$-3$$
=
-3
$$y_{1} = 5$$
=
$$5$$
=
5
Cramer's rule
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
We give the system of equations to the canonical form
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 2 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\11\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\3 & 4\end{matrix}\right] \right)} = -2$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 2\\11 & 4\end{matrix}\right] \right)}}{2} = -3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 7\\3 & 11\end{matrix}\right] \right)}}{2} = 5$$
$$3 x + 4 y = 11$$
We give the system of equations to the canonical form
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 2 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\11\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\3 & 4\end{matrix}\right] \right)} = -2$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 2\\11 & 4\end{matrix}\right] \right)}}{2} = -3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 7\\3 & 11\end{matrix}\right] \right)}}{2} = 5$$
Gaussian elimination
Given the system of equations
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
We give the system of equations to the canonical form
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 2 & 7\\3 & 4 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 2 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - 2 \cdot 3 & 11 - 3 \cdot 7\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 2 & 7\\0 & -2 & -10\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\-2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -2 & -10\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \left(-1\right) 0 & 2 - - -2 & 7 - - -10\end{matrix}\right] = \left[\begin{matrix}1 & 0 & -3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & -3\\0 & -2 & -10\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + 3 = 0$$
$$10 - 2 x_{2} = 0$$
We get the answer:
$$x_{1} = -3$$
$$x_{2} = 5$$
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
We give the system of equations to the canonical form
$$x + 2 y = 7$$
$$3 x + 4 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 2 & 7\\3 & 4 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 2 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - 2 \cdot 3 & 11 - 3 \cdot 7\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 2 & 7\\0 & -2 & -10\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\-2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -2 & -10\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \left(-1\right) 0 & 2 - - -2 & 7 - - -10\end{matrix}\right] = \left[\begin{matrix}1 & 0 & -3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & -3\\0 & -2 & -10\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} + 3 = 0$$
$$10 - 2 x_{2} = 0$$
We get the answer:
$$x_{1} = -3$$
$$x_{2} = 5$$