Mister Exam
2x+5y=11; 3x--2y=-12
The solution
You have entered
[src]
2*x + 5*y = 11
$$2 x + 5 y = 11$$
3*x + 2*y = -12
$$3 x + 2 y = -12$$
3*x + 2*y = -12
Detail solution
Given the system of equations
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
Let's express from equation 1 x
$$2 x + 5 y = 11$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 11 - 5 y$$
$$2 x = 11 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{11 - 5 y}{2}$$
$$x = \frac{11}{2} - \frac{5 y}{2}$$
Let's try the obtained element x to 2-th equation
$$3 x + 2 y = -12$$
We get:
$$2 y + 3 \left(\frac{11}{2} - \frac{5 y}{2}\right) = -12$$
$$\frac{33}{2} - \frac{11 y}{2} = -12$$
We move the free summand 33/2 from the left part to the right part performing the sign change
$$- \frac{11 y}{2} = - \frac{33}{2} - 12$$
$$- \frac{11 y}{2} = - \frac{57}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{11}{2} y}{- \frac{11}{2}} = - \frac{57}{\left(- \frac{11}{2}\right) 2}$$
$$y = \frac{57}{11}$$
Because
$$x = \frac{11}{2} - \frac{5 y}{2}$$
then
$$x = \frac{11}{2} - \frac{285}{22}$$
$$x = - \frac{82}{11}$$
The answer:
$$x = - \frac{82}{11}$$
$$y = \frac{57}{11}$$
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
Let's express from equation 1 x
$$2 x + 5 y = 11$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 11 - 5 y$$
$$2 x = 11 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{11 - 5 y}{2}$$
$$x = \frac{11}{2} - \frac{5 y}{2}$$
Let's try the obtained element x to 2-th equation
$$3 x + 2 y = -12$$
We get:
$$2 y + 3 \left(\frac{11}{2} - \frac{5 y}{2}\right) = -12$$
$$\frac{33}{2} - \frac{11 y}{2} = -12$$
We move the free summand 33/2 from the left part to the right part performing the sign change
$$- \frac{11 y}{2} = - \frac{33}{2} - 12$$
$$- \frac{11 y}{2} = - \frac{57}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{11}{2} y}{- \frac{11}{2}} = - \frac{57}{\left(- \frac{11}{2}\right) 2}$$
$$y = \frac{57}{11}$$
Because
$$x = \frac{11}{2} - \frac{5 y}{2}$$
then
$$x = \frac{11}{2} - \frac{285}{22}$$
$$x = - \frac{82}{11}$$
The answer:
$$x = - \frac{82}{11}$$
$$y = \frac{57}{11}$$
Rapid solution
$$x_{1} = - \frac{82}{11}$$
=
$$- \frac{82}{11}$$
=
$$y_{1} = \frac{57}{11}$$
=
$$\frac{57}{11}$$
=
=
$$- \frac{82}{11}$$
=
-7.45454545454545
$$y_{1} = \frac{57}{11}$$
=
$$\frac{57}{11}$$
=
5.18181818181818
Cramer's rule
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
We give the system of equations to the canonical form
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 5 x_{2}\\3 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}11\\-12\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 5\\3 & 2\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}11 & 5\\-12 & 2\end{matrix}\right] \right)}}{11} = - \frac{82}{11}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 11\\3 & -12\end{matrix}\right] \right)}}{11} = \frac{57}{11}$$
$$3 x + 2 y = -12$$
We give the system of equations to the canonical form
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + 5 x_{2}\\3 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}11\\-12\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 5\\3 & 2\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}11 & 5\\-12 & 2\end{matrix}\right] \right)}}{11} = - \frac{82}{11}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 11\\3 & -12\end{matrix}\right] \right)}}{11} = \frac{57}{11}$$
Gaussian elimination
Given the system of equations
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
We give the system of equations to the canonical form
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 5 & 11\\3 & 2 & -12\end{matrix}\right]$$
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
$$\left[\begin{matrix}2 & 5 & 11\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 2 - \frac{3 \cdot 5}{2} & - \frac{3 \cdot 11}{2} - 12\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 5 & 11\\0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\- \frac{11}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-10\right) 0}{11} & 5 - - -5 & 11 - - \frac{-285}{11}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & - \frac{164}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & - \frac{164}{11}\\0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} + \frac{164}{11} = 0$$
$$\frac{57}{2} - \frac{11 x_{2}}{2} = 0$$
We get the answer:
$$x_{1} = - \frac{82}{11}$$
$$x_{2} = \frac{57}{11}$$
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
We give the system of equations to the canonical form
$$2 x + 5 y = 11$$
$$3 x + 2 y = -12$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 5 & 11\\3 & 2 & -12\end{matrix}\right]$$
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
$$\left[\begin{matrix}2 & 5 & 11\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{2 \cdot 3}{2} & 2 - \frac{3 \cdot 5}{2} & - \frac{3 \cdot 11}{2} - 12\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 5 & 11\\0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\- \frac{11}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-10\right) 0}{11} & 5 - - -5 & 11 - - \frac{-285}{11}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & - \frac{164}{11}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & - \frac{164}{11}\\0 & - \frac{11}{2} & - \frac{57}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} + \frac{164}{11} = 0$$
$$\frac{57}{2} - \frac{11 x_{2}}{2} = 0$$
We get the answer:
$$x_{1} = - \frac{82}{11}$$
$$x_{2} = \frac{57}{11}$$
Numerical answer
[src]
x1 = -7.454545454545455 y1 = 5.181818181818182
x1 = -7.454545454545455 y1 = 5.181818181818182