Mister Exam
4x+2y=28; x+y=10
The solution
Detail solution
Given the system of equations
$$4 x + 2 y = 28$$
$$x + y = 10$$
Let's express from equation 1 x
$$4 x + 2 y = 28$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 28 - 2 y$$
$$4 x = 28 - 2 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{28 - 2 y}{4}$$
$$x = 7 - \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$x + y = 10$$
We get:
$$y + \left(7 - \frac{y}{2}\right) = 10$$
$$\frac{y}{2} + 7 = 10$$
We move the free summand 7 from the left part to the right part performing the sign change
$$\frac{y}{2} = -7 + 10$$
$$\frac{y}{2} = 3$$
Let's divide both parts of the equation by the multiplier of y
$$y = 6$$
Because
$$x = 7 - \frac{y}{2}$$
then
$$x = 7 - 3$$
$$x = 4$$
The answer:
$$x = 4$$
$$y = 6$$
$$4 x + 2 y = 28$$
$$x + y = 10$$
Let's express from equation 1 x
$$4 x + 2 y = 28$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 28 - 2 y$$
$$4 x = 28 - 2 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{28 - 2 y}{4}$$
$$x = 7 - \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$x + y = 10$$
We get:
$$y + \left(7 - \frac{y}{2}\right) = 10$$
$$\frac{y}{2} + 7 = 10$$
We move the free summand 7 from the left part to the right part performing the sign change
$$\frac{y}{2} = -7 + 10$$
$$\frac{y}{2} = 3$$
Let's divide both parts of the equation by the multiplier of y
/y\ |-| \2/ 3 --- = --- 1/2 1/2
$$y = 6$$
Because
$$x = 7 - \frac{y}{2}$$
then
$$x = 7 - 3$$
$$x = 4$$
The answer:
$$x = 4$$
$$y = 6$$
Rapid solution
$$x_{1} = 4$$
=
$$4$$
=
$$y_{1} = 6$$
=
$$6$$
=
=
$$4$$
=
4
$$y_{1} = 6$$
=
$$6$$
=
6
Gaussian elimination
Given the system of equations
$$4 x + 2 y = 28$$
$$x + y = 10$$
We give the system of equations to the canonical form
$$4 x + 2 y = 28$$
$$x + y = 10$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 2 & 28\\1 & 1 & 10\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 2 & 28\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{4}{4} & 1 - \frac{2}{4} & 10 - \frac{28}{4}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{2} & 3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 2 & 28\\0 & \frac{1}{2} & 3\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\\frac{1}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{2} & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 0 \cdot 4 & 2 - \frac{4}{2} & 28 - 3 \cdot 4\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 16\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 16\\0 & \frac{1}{2} & 3\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 16 = 0$$
$$\frac{x_{2}}{2} - 3 = 0$$
We get the answer:
$$x_{1} = 4$$
$$x_{2} = 6$$
$$4 x + 2 y = 28$$
$$x + y = 10$$
We give the system of equations to the canonical form
$$4 x + 2 y = 28$$
$$x + y = 10$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 2 & 28\\1 & 1 & 10\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 2 & 28\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{4}{4} & 1 - \frac{2}{4} & 10 - \frac{28}{4}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{1}{2} & 3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 2 & 28\\0 & \frac{1}{2} & 3\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\\frac{1}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{1}{2} & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - 0 \cdot 4 & 2 - \frac{4}{2} & 28 - 3 \cdot 4\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 16\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 16\\0 & \frac{1}{2} & 3\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 16 = 0$$
$$\frac{x_{2}}{2} - 3 = 0$$
We get the answer:
$$x_{1} = 4$$
$$x_{2} = 6$$
Cramer's rule
$$4 x + 2 y = 28$$
$$x + y = 10$$
We give the system of equations to the canonical form
$$4 x + 2 y = 28$$
$$x + y = 10$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 2 x_{2}\\x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}28\\10\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 2\\1 & 1\end{matrix}\right] \right)} = 2$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}28 & 2\\10 & 1\end{matrix}\right] \right)}}{2} = 4$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 28\\1 & 10\end{matrix}\right] \right)}}{2} = 6$$
$$x + y = 10$$
We give the system of equations to the canonical form
$$4 x + 2 y = 28$$
$$x + y = 10$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + 2 x_{2}\\x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}28\\10\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 2\\1 & 1\end{matrix}\right] \right)} = 2$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}28 & 2\\10 & 1\end{matrix}\right] \right)}}{2} = 4$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 28\\1 & 10\end{matrix}\right] \right)}}{2} = 6$$