Mister Exam
x+y=500; 1,3x+1,2y=630
The solution
You have entered
[src]
x + y = 500
$$x + y = 500$$
13*x 6*y ---- + --- = 630 10 5
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
13*x/10 + 6*y/5 = 630
Detail solution
Given the system of equations
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
Let's express from equation 1 x
$$x + y = 500$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = 500 - y$$
$$x = 500 - y$$
Let's try the obtained element x to 2-th equation
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
We get:
$$\frac{6 y}{5} + \frac{13 \left(500 - y\right)}{10} = 630$$
$$650 - \frac{y}{10} = 630$$
We move the free summand 650 from the left part to the right part performing the sign change
$$- \frac{y}{10} = -650 + 630$$
$$- \frac{y}{10} = -20$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{1}{10} y}{- \frac{1}{10}} = - \frac{20}{- \frac{1}{10}}$$
$$y = 200$$
Because
$$x = 500 - y$$
then
$$x = 500 - 200$$
$$x = 300$$
The answer:
$$x = 300$$
$$y = 200$$
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
Let's express from equation 1 x
$$x + y = 500$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = 500 - y$$
$$x = 500 - y$$
Let's try the obtained element x to 2-th equation
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
We get:
$$\frac{6 y}{5} + \frac{13 \left(500 - y\right)}{10} = 630$$
$$650 - \frac{y}{10} = 630$$
We move the free summand 650 from the left part to the right part performing the sign change
$$- \frac{y}{10} = -650 + 630$$
$$- \frac{y}{10} = -20$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{1}{10} y}{- \frac{1}{10}} = - \frac{20}{- \frac{1}{10}}$$
$$y = 200$$
Because
$$x = 500 - y$$
then
$$x = 500 - 200$$
$$x = 300$$
The answer:
$$x = 300$$
$$y = 200$$
Rapid solution
$$x_{1} = 300$$
=
$$300$$
=
$$y_{1} = 200$$
=
$$200$$
=
=
$$300$$
=
300
$$y_{1} = 200$$
=
$$200$$
=
200
Cramer's rule
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
We give the system of equations to the canonical form
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + x_{2}\\\frac{13 x_{1}}{10} + \frac{6 x_{2}}{5}\end{matrix}\right] = \left[\begin{matrix}500\\630\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 1\\\frac{13}{10} & \frac{6}{5}\end{matrix}\right] \right)} = - \frac{1}{10}$$
, 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 )
$$x_{1} = - 10 \operatorname{det}{\left(\left[\begin{matrix}500 & 1\\630 & \frac{6}{5}\end{matrix}\right] \right)} = 300$$
$$x_{2} = - 10 \operatorname{det}{\left(\left[\begin{matrix}1 & 500\\\frac{13}{10} & 630\end{matrix}\right] \right)} = 200$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
We give the system of equations to the canonical form
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + x_{2}\\\frac{13 x_{1}}{10} + \frac{6 x_{2}}{5}\end{matrix}\right] = \left[\begin{matrix}500\\630\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 1\\\frac{13}{10} & \frac{6}{5}\end{matrix}\right] \right)} = - \frac{1}{10}$$
, 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 )
$$x_{1} = - 10 \operatorname{det}{\left(\left[\begin{matrix}500 & 1\\630 & \frac{6}{5}\end{matrix}\right] \right)} = 300$$
$$x_{2} = - 10 \operatorname{det}{\left(\left[\begin{matrix}1 & 500\\\frac{13}{10} & 630\end{matrix}\right] \right)} = 200$$
Gaussian elimination
Given the system of equations
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
We give the system of equations to the canonical form
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 1 & 500\\\frac{13}{10} & \frac{6}{5} & 630\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\\frac{13}{10}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 1 & 500\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{\left(-1\right) 13}{10} + \frac{13}{10} & \frac{\left(-1\right) 13}{10} + \frac{6}{5} & 630 - \frac{13 \cdot 500}{10}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{1}{10} & -20\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & 500\\0 & - \frac{1}{10} & -20\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{1}{10}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{1}{10} & -20\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \left(-10\right) 0 & 1 - - -1 & 500 - - -200\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 300\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 300\\0 & - \frac{1}{10} & -20\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 300 = 0$$
$$20 - \frac{x_{2}}{10} = 0$$
We get the answer:
$$x_{1} = 300$$
$$x_{2} = 200$$
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
We give the system of equations to the canonical form
$$x + y = 500$$
$$\frac{13 x}{10} + \frac{6 y}{5} = 630$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 1 & 500\\\frac{13}{10} & \frac{6}{5} & 630\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\\frac{13}{10}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 1 & 500\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{\left(-1\right) 13}{10} + \frac{13}{10} & \frac{\left(-1\right) 13}{10} + \frac{6}{5} & 630 - \frac{13 \cdot 500}{10}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{1}{10} & -20\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 1 & 500\\0 & - \frac{1}{10} & -20\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{1}{10}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{1}{10} & -20\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \left(-10\right) 0 & 1 - - -1 & 500 - - -200\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 300\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 300\\0 & - \frac{1}{10} & -20\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 300 = 0$$
$$20 - \frac{x_{2}}{10} = 0$$
We get the answer:
$$x_{1} = 300$$
$$x_{2} = 200$$