Mister Exam
3x-2y=3; 3x+8y=-27
The solution
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3*x - 2*y = 3
$$3 x - 2 y = 3$$
3*x + 8*y = -27
$$3 x + 8 y = -27$$
3*x + 8*y = -27
Detail solution
Given the system of equations
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
Let's express from equation 1 x
$$3 x - 2 y = 3$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 2 y + 3$$
$$3 x = 2 y + 3$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{2 y + 3}{3}$$
$$x = \frac{2 y}{3} + 1$$
Let's try the obtained element x to 2-th equation
$$3 x + 8 y = -27$$
We get:
$$8 y + 3 \left(\frac{2 y}{3} + 1\right) = -27$$
$$10 y + 3 = -27$$
We move the free summand 3 from the left part to the right part performing the sign change
$$10 y = -27 - 3$$
$$10 y = -30$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{10 y}{10} = - \frac{30}{10}$$
$$y = -3$$
Because
$$x = \frac{2 y}{3} + 1$$
then
$$x = \frac{\left(-3\right) 2}{3} + 1$$
$$x = -1$$
The answer:
$$x = -1$$
$$y = -3$$
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
Let's express from equation 1 x
$$3 x - 2 y = 3$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 2 y + 3$$
$$3 x = 2 y + 3$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{2 y + 3}{3}$$
$$x = \frac{2 y}{3} + 1$$
Let's try the obtained element x to 2-th equation
$$3 x + 8 y = -27$$
We get:
$$8 y + 3 \left(\frac{2 y}{3} + 1\right) = -27$$
$$10 y + 3 = -27$$
We move the free summand 3 from the left part to the right part performing the sign change
$$10 y = -27 - 3$$
$$10 y = -30$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{10 y}{10} = - \frac{30}{10}$$
$$y = -3$$
Because
$$x = \frac{2 y}{3} + 1$$
then
$$x = \frac{\left(-3\right) 2}{3} + 1$$
$$x = -1$$
The answer:
$$x = -1$$
$$y = -3$$
Rapid solution
$$x_{1} = -1$$
=
$$-1$$
=
$$y_{1} = -3$$
=
$$-3$$
=
=
$$-1$$
=
-1
$$y_{1} = -3$$
=
$$-3$$
=
-3
Gaussian elimination
Given the system of equations
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
We give the system of equations to the canonical form
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 3\\3 & 8 & -27\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 8 - -2 & -27 + \left(-1\right) 3\end{matrix}\right] = \left[\begin{matrix}0 & 10 & -30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 3\\0 & 10 & -30\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\10\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 10 & -30\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-1\right) 0}{5} & -2 - \frac{\left(-1\right) 10}{5} & 3 - - -6\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -3\\0 & 10 & -30\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} + 3 = 0$$
$$10 x_{2} + 30 = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = -3$$
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
We give the system of equations to the canonical form
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -2 & 3\\3 & 8 & -27\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -2 & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 8 - -2 & -27 + \left(-1\right) 3\end{matrix}\right] = \left[\begin{matrix}0 & 10 & -30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -2 & 3\\0 & 10 & -30\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-2\\10\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 10 & -30\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-1\right) 0}{5} & -2 - \frac{\left(-1\right) 10}{5} & 3 - - -6\end{matrix}\right] = \left[\begin{matrix}3 & 0 & -3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & -3\\0 & 10 & -30\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} + 3 = 0$$
$$10 x_{2} + 30 = 0$$
We get the answer:
$$x_{1} = -1$$
$$x_{2} = -3$$
Cramer's rule
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
We give the system of equations to the canonical form
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2}\\3 x_{1} + 8 x_{2}\end{matrix}\right] = \left[\begin{matrix}3\\-27\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2\\3 & 8\end{matrix}\right] \right)} = 30$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -2\\-27 & 8\end{matrix}\right] \right)}}{30} = -1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 3\\3 & -27\end{matrix}\right] \right)}}{30} = -3$$
$$3 x + 8 y = -27$$
We give the system of equations to the canonical form
$$3 x - 2 y = 3$$
$$3 x + 8 y = -27$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} - 2 x_{2}\\3 x_{1} + 8 x_{2}\end{matrix}\right] = \left[\begin{matrix}3\\-27\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & -2\\3 & 8\end{matrix}\right] \right)} = 30$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & -2\\-27 & 8\end{matrix}\right] \right)}}{30} = -1$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 3\\3 & -27\end{matrix}\right] \right)}}{30} = -3$$