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System of equations solver/ Х+2у=7; 3х+4у=11

Х+2у=7; 3х+4у=11

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x + 2*y = 7
x+2y=7x + 2 y = 7 
3*x + 4*y = 11
3x+4y=113 x + 4 y = 11 
3*x + 4*y = 11
Detail solution
Given the system of equations
x+2y=7x + 2 y = 7
3x+4y=113 x + 4 y = 11

Let's express from equation 1 x
x+2y=7x + 2 y = 7
Let's move the summand with the variable y from the left part to the right part performing the sign change
x=7−2yx = 7 - 2 y
x=7−2yx = 7 - 2 y
Let's try the obtained element x to 2-th equation
3x+4y=113 x + 4 y = 11
We get:
4y+3(7−2y)=114 y + 3 \left(7 - 2 y\right) = 11
21−2y=1121 - 2 y = 11
We move the free summand 21 from the left part to the right part performing the sign change
−2y=−21+11- 2 y = -21 + 11
−2y=−10- 2 y = -10
Let's divide both parts of the equation by the multiplier of y
(−1)2y−2=−10−2\frac{\left(-1\right) 2 y}{-2} = - \frac{10}{-2}
y=5y = 5
Because
x=7−2yx = 7 - 2 y
then
x=7−10x = 7 - 10
x=−3x = -3

The answer:
x=−3x = -3
y=5y = 5
Rapid solution
x1=−3x_{1} = -3
=
−3-3
=
-3

y1=5y_{1} = 5
=
55
=
5
Cramer's rule
x+2y=7x + 2 y = 7
3x+4y=113 x + 4 y = 11

We give the system of equations to the canonical form
x+2y=7x + 2 y = 7
3x+4y=113 x + 4 y = 11
Rewrite the system of linear equations as the matrix form
[x1+2x23x1+4x2]=[711]\left[\begin{matrix}x_{1} + 2 x_{2}\\3 x_{1} + 4 x_{2}\end{matrix}\right] = \left[\begin{matrix}7\\11\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=det⁡([1234])=−2A = \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\3 & 4\end{matrix}\right] \right)} = -2
, 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 )
x1=−det⁡([72114])2=−3x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}7 & 2\\11 & 4\end{matrix}\right] \right)}}{2} = -3
x2=−det⁡([17311])2=5x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & 7\\3 & 11\end{matrix}\right] \right)}}{2} = 5
Gaussian elimination
Given the system of equations
x+2y=7x + 2 y = 7
3x+4y=113 x + 4 y = 11

We give the system of equations to the canonical form
x+2y=7x + 2 y = 7
3x+4y=113 x + 4 y = 11
Rewrite the system of linear equations as the matrix form
[1273411]\left[\begin{matrix}1 & 2 & 7\\3 & 4 & 11\end{matrix}\right]
In 1 -th column
[13]\left[\begin{matrix}1\\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
[127]\left[\begin{matrix}1 & 2 & 7\end{matrix}\right]
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
[(−1)3+34−2⋅311−3⋅7]=[0−2−10]\left[\begin{matrix}\left(-1\right) 3 + 3 & 4 - 2 \cdot 3 & 11 - 3 \cdot 7\end{matrix}\right] = \left[\begin{matrix}0 & -2 & -10\end{matrix}\right]
you get
[1270−2−10]\left[\begin{matrix}1 & 2 & 7\\0 & -2 & -10\end{matrix}\right]
In 2 -th column
[2−2]\left[\begin{matrix}2\\-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
[0−2−10]\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:
[1−(−1)02−−−27−−−10]=[10−3]\left[\begin{matrix}1 - \left(-1\right) 0 & 2 - - -2 & 7 - - -10\end{matrix}\right] = \left[\begin{matrix}1 & 0 & -3\end{matrix}\right]
you get
[10−30−2−10]\left[\begin{matrix}1 & 0 & -3\\0 & -2 & -10\end{matrix}\right]

It is almost ready, all we have to do is to find variables, solving the elementary equations:
x1+3=0x_{1} + 3 = 0
10−2x2=010 - 2 x_{2} = 0
We get the answer:
x1=−3x_{1} = -3
x2=5x_{2} = 5
Numerical answer [src] 
x1 = -3.0
y1 = 5.0
x1 = -3.0
y1 = 5.0
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