Mister Exam
y=2x-1; -2x+3y=9
The solution
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y = 2*x - 1
$$y = 2 x - 1$$
-2*x + 3*y = 9
$$- 2 x + 3 y = 9$$
-2*x + 3*y = 9
Detail solution
Given the system of equations
$$y = 2 x - 1$$
$$- 2 x + 3 y = 9$$
Let's express from equation 1 x
$$y = 2 x - 1$$
Let's move the summand with the variable x from the right part to the left part performing the sign change
$$- 2 x + y = -1$$
$$- 2 x + y = -1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$- 2 x = - y - 1$$
$$- 2 x = - y - 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\left(-1\right) 2 x}{-2} = \frac{- y - 1}{-2}$$
$$x = \frac{y}{2} + \frac{1}{2}$$
Let's try the obtained element x to 2-th equation
$$- 2 x + 3 y = 9$$
We get:
$$3 y - 2 \left(\frac{y}{2} + \frac{1}{2}\right) = 9$$
$$2 y - 1 = 9$$
We move the free summand -1 from the left part to the right part performing the sign change
$$2 y = 1 + 9$$
$$2 y = 10$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{2 y}{2} = \frac{10}{2}$$
$$y = 5$$
Because
$$x = \frac{y}{2} + \frac{1}{2}$$
then
$$x = \frac{1}{2} + \frac{5}{2}$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = 5$$
$$y = 2 x - 1$$
$$- 2 x + 3 y = 9$$
Let's express from equation 1 x
$$y = 2 x - 1$$
Let's move the summand with the variable x from the right part to the left part performing the sign change
$$- 2 x + y = -1$$
$$- 2 x + y = -1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$- 2 x = - y - 1$$
$$- 2 x = - y - 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\left(-1\right) 2 x}{-2} = \frac{- y - 1}{-2}$$
$$x = \frac{y}{2} + \frac{1}{2}$$
Let's try the obtained element x to 2-th equation
$$- 2 x + 3 y = 9$$
We get:
$$3 y - 2 \left(\frac{y}{2} + \frac{1}{2}\right) = 9$$
$$2 y - 1 = 9$$
We move the free summand -1 from the left part to the right part performing the sign change
$$2 y = 1 + 9$$
$$2 y = 10$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{2 y}{2} = \frac{10}{2}$$
$$y = 5$$
Because
$$x = \frac{y}{2} + \frac{1}{2}$$
then
$$x = \frac{1}{2} + \frac{5}{2}$$
$$x = 3$$
The answer:
$$x = 3$$
$$y = 5$$
Rapid solution
$$x_{1} = 3$$
=
$$3$$
=
$$y_{1} = 5$$
=
$$5$$
=
=
$$3$$
=
3
$$y_{1} = 5$$
=
$$5$$
=
5
Cramer's rule
$$y = 2 x - 1$$
$$- 2 x + 3 y = 9$$
We give the system of equations to the canonical form
$$- 2 x + y = -1$$
$$- 2 x + 3 y = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- 2 x_{1} + x_{2}\\- 2 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}-1\\9\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-2 & 1\\-2 & 3\end{matrix}\right] \right)} = -4$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-1 & 1\\9 & 3\end{matrix}\right] \right)}}{4} = 3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-2 & -1\\-2 & 9\end{matrix}\right] \right)}}{4} = 5$$
$$- 2 x + 3 y = 9$$
We give the system of equations to the canonical form
$$- 2 x + y = -1$$
$$- 2 x + 3 y = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- 2 x_{1} + x_{2}\\- 2 x_{1} + 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}-1\\9\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-2 & 1\\-2 & 3\end{matrix}\right] \right)} = -4$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-1 & 1\\9 & 3\end{matrix}\right] \right)}}{4} = 3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-2 & -1\\-2 & 9\end{matrix}\right] \right)}}{4} = 5$$
Gaussian elimination
Given the system of equations
$$y = 2 x - 1$$
$$- 2 x + 3 y = 9$$
We give the system of equations to the canonical form
$$- 2 x + y = -1$$
$$- 2 x + 3 y = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-2 & 1 & -1\\-2 & 3 & 9\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-2\\-2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-2 & 1 & -1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-2 - -2 & -1 + 3 & 9 - -1\end{matrix}\right] = \left[\begin{matrix}0 & 2 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-2 & 1 & -1\\0 & 2 & 10\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 2 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-2 - \frac{0}{2} & 1 - \frac{2}{2} & - \frac{10}{2} - 1\end{matrix}\right] = \left[\begin{matrix}-2 & 0 & -6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-2 & 0 & -6\\0 & 2 & 10\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$6 - 2 x_{1} = 0$$
$$2 x_{2} - 10 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 5$$
$$y = 2 x - 1$$
$$- 2 x + 3 y = 9$$
We give the system of equations to the canonical form
$$- 2 x + y = -1$$
$$- 2 x + 3 y = 9$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-2 & 1 & -1\\-2 & 3 & 9\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-2\\-2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-2 & 1 & -1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-2 - -2 & -1 + 3 & 9 - -1\end{matrix}\right] = \left[\begin{matrix}0 & 2 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-2 & 1 & -1\\0 & 2 & 10\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 2 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-2 - \frac{0}{2} & 1 - \frac{2}{2} & - \frac{10}{2} - 1\end{matrix}\right] = \left[\begin{matrix}-2 & 0 & -6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-2 & 0 & -6\\0 & 2 & 10\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$6 - 2 x_{1} = 0$$
$$2 x_{2} - 10 = 0$$
We get the answer:
$$x_{1} = 3$$
$$x_{2} = 5$$