Mister Exam
2y+x=-8; 5x-4=14
The solution
Detail solution
Given the system of equations
$$x + 2 y = -8$$
$$5 x - 4 = 14$$
Let's express from equation 1 x
$$x + 2 y = -8$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = - 2 y - 8$$
$$x = - 2 y - 8$$
Let's try the obtained element x to 2-th equation
$$5 x - 4 = 14$$
We get:
$$5 \left(- 2 y - 8\right) - 4 = 14$$
$$- 10 y - 44 = 14$$
We move the free summand -44 from the left part to the right part performing the sign change
$$- 10 y = 14 + 44$$
$$- 10 y = 58$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 10 y}{-10} = \frac{58}{-10}$$
$$y = - \frac{29}{5}$$
Because
$$x = - 2 y - 8$$
then
$$x = -8 - - \frac{58}{5}$$
$$x = \frac{18}{5}$$
The answer:
$$x = \frac{18}{5}$$
$$y = - \frac{29}{5}$$
$$x + 2 y = -8$$
$$5 x - 4 = 14$$
Let's express from equation 1 x
$$x + 2 y = -8$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x = - 2 y - 8$$
$$x = - 2 y - 8$$
Let's try the obtained element x to 2-th equation
$$5 x - 4 = 14$$
We get:
$$5 \left(- 2 y - 8\right) - 4 = 14$$
$$- 10 y - 44 = 14$$
We move the free summand -44 from the left part to the right part performing the sign change
$$- 10 y = 14 + 44$$
$$- 10 y = 58$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 10 y}{-10} = \frac{58}{-10}$$
$$y = - \frac{29}{5}$$
Because
$$x = - 2 y - 8$$
then
$$x = -8 - - \frac{58}{5}$$
$$x = \frac{18}{5}$$
The answer:
$$x = \frac{18}{5}$$
$$y = - \frac{29}{5}$$
Rapid solution
$$x_{1} = \frac{18}{5}$$
=
$$\frac{18}{5}$$
=
$$y_{1} = - \frac{29}{5}$$
=
$$- \frac{29}{5}$$
=
=
$$\frac{18}{5}$$
=
3.6
$$y_{1} = - \frac{29}{5}$$
=
$$- \frac{29}{5}$$
=
-5.8
Gaussian elimination
Given the system of equations
$$x + 2 y = -8$$
$$5 x - 4 = 14$$
We give the system of equations to the canonical form
$$x + 2 y = -8$$
$$5 x = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 2 & -8\\5 & 0 & 18\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}5 & 0 & 18\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{5}{5} & 2 - \frac{0}{5} & -8 - \frac{18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & 2 & - \frac{58}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & 2 & - \frac{58}{5}\\5 & 0 & 18\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{2} + \frac{58}{5} = 0$$
$$5 x_{1} - 18 = 0$$
We get the answer:
$$x_{2} = - \frac{29}{5}$$
$$x_{1} = \frac{18}{5}$$
$$x + 2 y = -8$$
$$5 x - 4 = 14$$
We give the system of equations to the canonical form
$$x + 2 y = -8$$
$$5 x = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 2 & -8\\5 & 0 & 18\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}5 & 0 & 18\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{5}{5} & 2 - \frac{0}{5} & -8 - \frac{18}{5}\end{matrix}\right] = \left[\begin{matrix}0 & 2 & - \frac{58}{5}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}0 & 2 & - \frac{58}{5}\\5 & 0 & 18\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{2} + \frac{58}{5} = 0$$
$$5 x_{1} - 18 = 0$$
We get the answer:
$$x_{2} = - \frac{29}{5}$$
$$x_{1} = \frac{18}{5}$$
Cramer's rule
$$x + 2 y = -8$$
$$5 x - 4 = 14$$
We give the system of equations to the canonical form
$$x + 2 y = -8$$
$$5 x = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 2 x_{2}\\5 x_{1} + 0 x_{2}\end{matrix}\right] = \left[\begin{matrix}-8\\18\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\5 & 0\end{matrix}\right] \right)} = -10$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-8 & 2\\18 & 0\end{matrix}\right] \right)}}{10} = \frac{18}{5}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -8\\5 & 18\end{matrix}\right] \right)}}{10} = - \frac{29}{5}$$
$$5 x - 4 = 14$$
We give the system of equations to the canonical form
$$x + 2 y = -8$$
$$5 x = 18$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} + 2 x_{2}\\5 x_{1} + 0 x_{2}\end{matrix}\right] = \left[\begin{matrix}-8\\18\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\5 & 0\end{matrix}\right] \right)} = -10$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-8 & 2\\18 & 0\end{matrix}\right] \right)}}{10} = \frac{18}{5}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -8\\5 & 18\end{matrix}\right] \right)}}{10} = - \frac{29}{5}$$