Mister Exam
2x+5y=-8; 2+3y=-4
The solution
Detail solution
Given the system of equations
$$2 x + 5 y = -8$$
$$3 y + 2 = -4$$
Let's express from equation 1 x
$$2 x + 5 y = -8$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = - 5 y - 8$$
$$2 x = - 5 y - 8$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{- 5 y - 8}{2}$$
$$x = - \frac{5 y}{2} - 4$$
Let's try the obtained element x to 2-th equation
$$3 y + 2 = -4$$
We get:
$$3 y + 2 = -4$$
$$3 y + 2 = -4$$
We move the free summand 2 from the left part to the right part performing the sign change
$$3 y = -4 - 2$$
$$3 y = -6$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{3 y}{3} = - \frac{6}{3}$$
$$y = -2$$
Because
$$x = - \frac{5 y}{2} - 4$$
then
$$x = -4 - -5$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = -2$$
$$2 x + 5 y = -8$$
$$3 y + 2 = -4$$
Let's express from equation 1 x
$$2 x + 5 y = -8$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = - 5 y - 8$$
$$2 x = - 5 y - 8$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{- 5 y - 8}{2}$$
$$x = - \frac{5 y}{2} - 4$$
Let's try the obtained element x to 2-th equation
$$3 y + 2 = -4$$
We get:
$$3 y + 2 = -4$$
$$3 y + 2 = -4$$
We move the free summand 2 from the left part to the right part performing the sign change
$$3 y = -4 - 2$$
$$3 y = -6$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{3 y}{3} = - \frac{6}{3}$$
$$y = -2$$
Because
$$x = - \frac{5 y}{2} - 4$$
then
$$x = -4 - -5$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = -2$$
Rapid solution
$$x_{1} = 1$$
=
$$1$$
=
$$y_{1} = -2$$
=
$$-2$$
=
=
$$1$$
=
1
$$y_{1} = -2$$
=
$$-2$$
=
-2
Cramer's rule
$$2 x + 5 y = -8$$
$$3 y + 2 = -4$$
We give the system of equations to the canonical form
$$2 x + 5 y = -8$$
$$3 y = -6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 5 x_{2}\\0 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}-8\\-6\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 5\\0 & 3\end{matrix}\right] \right)} = 6$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-8 & 5\\-6 & 3\end{matrix}\right] \right)}}{6} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -8\\0 & -6\end{matrix}\right] \right)}}{6} = -2$$
$$3 y + 2 = -4$$
We give the system of equations to the canonical form
$$2 x + 5 y = -8$$
$$3 y = -6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 5 x_{2}\\0 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}-8\\-6\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 5\\0 & 3\end{matrix}\right] \right)} = 6$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-8 & 5\\-6 & 3\end{matrix}\right] \right)}}{6} = 1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & -8\\0 & -6\end{matrix}\right] \right)}}{6} = -2$$
Gaussian elimination
Given the system of equations
$$2 x + 5 y = -8$$
$$3 y + 2 = -4$$
We give the system of equations to the canonical form
$$2 x + 5 y = -8$$
$$3 y = -6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 5 & -8\\0 & 3 & -6\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 5 & -8\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}5\\3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & -6\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{0 \cdot 5}{3} & 5 - \frac{3 \cdot 5}{3} & -8 - - 10\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 2\\0 & 3 & -6\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 2 = 0$$
$$3 x_{2} + 6 = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -2$$
$$2 x + 5 y = -8$$
$$3 y + 2 = -4$$
We give the system of equations to the canonical form
$$2 x + 5 y = -8$$
$$3 y = -6$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 5 & -8\\0 & 3 & -6\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 5 & -8\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}5\\3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & -6\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{0 \cdot 5}{3} & 5 - \frac{3 \cdot 5}{3} & -8 - - 10\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 2\\0 & 3 & -6\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 2 = 0$$
$$3 x_{2} + 6 = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -2$$