Mister Exam
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-
How to use it?
- 2x+y=-10; 6x+3y=-30
- 2х/9+у/4=11; 5х/12+у/3=19
- 2x+5y+4z+t=20; 1x+3y+2z+t=11; 2x+10y+9z+7t=40; 3x+8y+9z+2t=37
- x-2y+3z=6; 2x+3y-4z=20; 3x-2y-5z=6
- 2x+y
- Plot:
-
2x+y
- Canonical form:
- 2x+y
-
Identical expressions
- 2x+y=- ten ; 6x+3y=- thirty
- 2x plus y equally minus 10; 6x plus 3y equally minus 30
- 2x plus y equally minus ten ; 6x plus 3y equally minus thirty
-
Similar expressions
- 2(3х-2y)+1=7x; 12(x+y)-15=7x+12y
- 2x+y=+10; 6x+3y=-30
- 2x-y=-10; 6x+3y=-30
- 2x+y=-10; 6x-3y=-30
- 2x+y=-10; 6x+3y=+30
2x+y=-10; 6x+3y=-30
The solution
You have entered
[src]
2*x + y = -10
$$2 x + y = -10$$
6*x + 3*y = -30
$$6 x + 3 y = -30$$
6*x + 3*y = -30
Detail solution
Given the system of equations
$$2 x + y = -10$$
$$6 x + 3 y = -30$$
Let's express from equation 1 x
$$2 x + y = -10$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = - y - 10$$
$$2 x = - y - 10$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{- y - 10}{2}$$
$$x = - \frac{y}{2} - 5$$
Let's try the obtained element x to 2-th equation
$$6 x + 3 y = -30$$
We get:
$$3 y + 6 \left(- \frac{y}{2} - 5\right) = -30$$
so
is always executed
$$2 x + y = -10$$
$$6 x + 3 y = -30$$
Let's express from equation 1 x
$$2 x + y = -10$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = - y - 10$$
$$2 x = - y - 10$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{- y - 10}{2}$$
$$x = - \frac{y}{2} - 5$$
Let's try the obtained element x to 2-th equation
$$6 x + 3 y = -30$$
We get:
$$3 y + 6 \left(- \frac{y}{2} - 5\right) = -30$$
so
is always executed
Rapid solution
$$x_{1} = - \frac{y}{2} - 5$$
=
$$- \frac{y}{2} - 5$$
=
=
$$- \frac{y}{2} - 5$$
=
-5 - 0.5*y
Gaussian elimination
Given the system of equations
$$2 x + y = -10$$
$$6 x + 3 y = -30$$
We give the system of equations to the canonical form
$$2 x + y = -10$$
$$6 x + 3 y = -30$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 1 & -10\\6 & 3 & -30\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & -10\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 2 \cdot 3 & \left(-1\right) 3 + 3 & -30 - - 30\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 1 & -10\\0 & 0 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & -10\end{matrix}\right]$$
,
and subtract it from other lines:
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} + x_{2} + 10 = 0$$
$$0 - 0 = 0$$
We get the answer:
$$x_{1} = - \frac{x_{2}}{2} - 5$$
where x2 - the free variables
$$2 x + y = -10$$
$$6 x + 3 y = -30$$
We give the system of equations to the canonical form
$$2 x + y = -10$$
$$6 x + 3 y = -30$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 1 & -10\\6 & 3 & -30\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & -10\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 2 \cdot 3 & \left(-1\right) 3 + 3 & -30 - - 30\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 1 & -10\\0 & 0 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & -10\end{matrix}\right]$$
,
and subtract it from other lines:
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} + x_{2} + 10 = 0$$
$$0 - 0 = 0$$
We get the answer:
$$x_{1} = - \frac{x_{2}}{2} - 5$$
where x2 - the free variables