Mister Exam
2x-4у=1; 3х+5у=-2
The solution
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2*x - 4*y = 1
$$2 x - 4 y = 1$$
3*x + 5*y = -2
$$3 x + 5 y = -2$$
3*x + 5*y = -2
Detail solution
Given the system of equations
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
Let's express from equation 1 x
$$2 x - 4 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 4 y + 1$$
$$2 x = 4 y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{4 y + 1}{2}$$
$$x = 2 y + \frac{1}{2}$$
Let's try the obtained element x to 2-th equation
$$3 x + 5 y = -2$$
We get:
$$5 y + 3 \left(2 y + \frac{1}{2}\right) = -2$$
$$11 y + \frac{3}{2} = -2$$
We move the free summand 3/2 from the left part to the right part performing the sign change
$$11 y = -2 - \frac{3}{2}$$
$$11 y = - \frac{7}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{11 y}{11} = - \frac{7}{2 \cdot 11}$$
$$y = - \frac{7}{22}$$
Because
$$x = 2 y + \frac{1}{2}$$
then
$$x = \frac{\left(-7\right) 2}{22} + \frac{1}{2}$$
$$x = - \frac{3}{22}$$
The answer:
$$x = - \frac{3}{22}$$
$$y = - \frac{7}{22}$$
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
Let's express from equation 1 x
$$2 x - 4 y = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 4 y + 1$$
$$2 x = 4 y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{4 y + 1}{2}$$
$$x = 2 y + \frac{1}{2}$$
Let's try the obtained element x to 2-th equation
$$3 x + 5 y = -2$$
We get:
$$5 y + 3 \left(2 y + \frac{1}{2}\right) = -2$$
$$11 y + \frac{3}{2} = -2$$
We move the free summand 3/2 from the left part to the right part performing the sign change
$$11 y = -2 - \frac{3}{2}$$
$$11 y = - \frac{7}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{11 y}{11} = - \frac{7}{2 \cdot 11}$$
$$y = - \frac{7}{22}$$
Because
$$x = 2 y + \frac{1}{2}$$
then
$$x = \frac{\left(-7\right) 2}{22} + \frac{1}{2}$$
$$x = - \frac{3}{22}$$
The answer:
$$x = - \frac{3}{22}$$
$$y = - \frac{7}{22}$$
Rapid solution
$$x_{1} = - \frac{3}{22}$$
=
$$- \frac{3}{22}$$
=
$$y_{1} = - \frac{7}{22}$$
=
$$- \frac{7}{22}$$
=
=
$$- \frac{3}{22}$$
=
-0.136363636363636
$$y_{1} = - \frac{7}{22}$$
=
$$- \frac{7}{22}$$
=
-0.318181818181818
Cramer's rule
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
We give the system of equations to the canonical form
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - 4 x_{2}\\3 x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\-2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -4\\3 & 5\end{matrix}\right] \right)} = 22$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -4\\-2 & 5\end{matrix}\right] \right)}}{22} = - \frac{3}{22}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 1\\3 & -2\end{matrix}\right] \right)}}{22} = - \frac{7}{22}$$
$$3 x + 5 y = -2$$
We give the system of equations to the canonical form
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} - 4 x_{2}\\3 x_{1} + 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\-2\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & -4\\3 & 5\end{matrix}\right] \right)} = 22$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -4\\-2 & 5\end{matrix}\right] \right)}}{22} = - \frac{3}{22}$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 1\\3 & -2\end{matrix}\right] \right)}}{22} = - \frac{7}{22}$$
Gaussian elimination
Given the system of equations
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
We give the system of equations to the canonical form
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -4 & 1\\3 & 5 & -2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -4 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 5 - - 6 & -2 + \frac{\left(-1\right) 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -4 & 1\\0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-4\\11\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{11} & -4 - \frac{\left(-4\right) 11}{11} & 1 - - \frac{-14}{11}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & - \frac{3}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & - \frac{3}{11}\\0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} + \frac{3}{11} = 0$$
$$11 x_{2} + \frac{7}{2} = 0$$
We get the answer:
$$x_{1} = - \frac{3}{22}$$
$$x_{2} = - \frac{7}{22}$$
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
We give the system of equations to the canonical form
$$2 x - 4 y = 1$$
$$3 x + 5 y = -2$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -4 & 1\\3 & 5 & -2\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -4 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 5 - - 6 & -2 + \frac{\left(-1\right) 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -4 & 1\\0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-4\\11\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{11} & -4 - \frac{\left(-4\right) 11}{11} & 1 - - \frac{-14}{11}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & - \frac{3}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & - \frac{3}{11}\\0 & 11 & - \frac{7}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} + \frac{3}{11} = 0$$
$$11 x_{2} + \frac{7}{2} = 0$$
We get the answer:
$$x_{1} = - \frac{3}{22}$$
$$x_{2} = - \frac{7}{22}$$
Numerical answer
[src]
x1 = -0.1363636363636364 y1 = -0.3181818181818182
x1 = -0.1363636363636364 y1 = -0.3181818181818182