Mister Exam
3х-у=15; 6х+у=12
The solution
Detail solution
Given the system of equations
$$3 x - y = 15$$
$$6 x + y = 12$$
Let's express from equation 1 x
$$3 x - y = 15$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = y + 15$$
$$3 x = y + 15$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{y + 15}{3}$$
$$x = \frac{y}{3} + 5$$
Let's try the obtained element x to 2-th equation
$$6 x + y = 12$$
We get:
$$y + 6 \left(\frac{y}{3} + 5\right) = 12$$
$$3 y + 30 = 12$$
We move the free summand 30 from the left part to the right part performing the sign change
$$3 y = -30 + 12$$
$$3 y = -18$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{3 y}{3} = - \frac{18}{3}$$
$$y = -6$$
Because
$$x = \frac{y}{3} + 5$$
then
$$x = \frac{-6}{3} + 5$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = -6$$
$$3 x - y = 15$$
$$6 x + y = 12$$
Let's express from equation 1 x
$$3 x - y = 15$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = y + 15$$
$$3 x = y + 15$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{y + 15}{3}$$
$$x = \frac{y}{3} + 5$$
Let's try the obtained element x to 2-th equation
$$6 x + y = 12$$
We get:
$$y + 6 \left(\frac{y}{3} + 5\right) = 12$$
$$3 y + 30 = 12$$
We move the free summand 30 from the left part to the right part performing the sign change
$$3 y = -30 + 12$$
$$3 y = -18$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{3 y}{3} = - \frac{18}{3}$$
$$y = -6$$
Because
$$x = \frac{y}{3} + 5$$
then
$$x = \frac{-6}{3} + 5$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = -6$$
Rapid solution
$$x_{1} = 3$$
=
$$3$$
=
$$y_{1} = -6$$
=
$$-6$$
=
=
$$3$$
=
3
$$y_{1} = -6$$
=
$$-6$$
=
-6
Gaussian elimination
Given the system of equations
$$3 x - y = 15$$
$$6 x + y = 12$$
We give the system of equations to the canonical form
$$3 x - y = 15$$
$$6 x + y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -1 & 15\\6 & 1 & 12\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -1 & 15\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 2 \cdot 3 & 1 - - 2 & 12 - 2 \cdot 15\end{matrix}\right] = \left[\begin{matrix}0 & 3 & -18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -1 & 15\\0 & 3 & -18\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & -18\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-1\right) 0}{3} & -1 - \frac{\left(-1\right) 3}{3} & 15 - - -6\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 9\\0 & 3 & -18\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 9 = 0$$
$$3 x_{2} + 18 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = -6$$
$$3 x - y = 15$$
$$6 x + y = 12$$
We give the system of equations to the canonical form
$$3 x - y = 15$$
$$6 x + y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -1 & 15\\6 & 1 & 12\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -1 & 15\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 2 \cdot 3 & 1 - - 2 & 12 - 2 \cdot 15\end{matrix}\right] = \left[\begin{matrix}0 & 3 & -18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -1 & 15\\0 & 3 & -18\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & -18\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-1\right) 0}{3} & -1 - \frac{\left(-1\right) 3}{3} & 15 - - -6\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 9\\0 & 3 & -18\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 9 = 0$$
$$3 x_{2} + 18 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = -6$$
Cramer's rule
$$3 x - y = 15$$
$$6 x + y = 12$$
We give the system of equations to the canonical form
$$3 x - y = 15$$
$$6 x + y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - x_{2}\\6 x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}15\\12\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -1\\6 & 1\end{matrix}\right] \right)} = 9$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}15 & -1\\12 & 1\end{matrix}\right] \right)}}{9} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 15\\6 & 12\end{matrix}\right] \right)}}{9} = -6$$
$$6 x + y = 12$$
We give the system of equations to the canonical form
$$3 x - y = 15$$
$$6 x + y = 12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - x_{2}\\6 x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}15\\12\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -1\\6 & 1\end{matrix}\right] \right)} = 9$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}15 & -1\\12 & 1\end{matrix}\right] \right)}}{9} = 3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 15\\6 & 12\end{matrix}\right] \right)}}{9} = -6$$