Mister Exam
Other calculators
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How to use it?
- 2x1+x2−x3+x4=1; 3x2−x4=1; x3−x4=0
- 2x1−3x2−x3=4; x1+x2−3x3=−3; 3x1+2x2+x3=2
-
16x+32y-212,479=0; 32x+128y-791,249=0
- log7(36−y^2)=log7(36−a^2x^2); x^2+y^2=2x+6y
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Identical expressions
- 2x one +x2−x3+x4= one ; 3x2−x4=1; x3−x4= zero
- 2x1 plus x2−x3 plus x4 equally 1; 3x2−x4 equally 1; x3−x4 equally 0
- 2x one plus x2−x3 plus x4 equally one ; 3x2−x4 equally 1; x3−x4 equally zero
- 2x1+x2−x3+x4=1; 3x2−x4=1; x3−x4=O
-
Similar expressions
- 2x1-x2−x3+x4=1; 3x2−x4=1; x3−x4=0
- 2x1+x2−x3-x4=1; 3x2−x4=1; x3−x4=0
2x1+x2−x3+x4=1; 3x2−x4=1; x3−x4=0
The solution
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2*x1 + x2 - x3 + x4 = 1
$$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$
3*x2 - x4 = 1
$$3 x_{2} - x_{4} = 1$$
x3 - x4 = 0
$$x_{3} - x_{4} = 0$$
or
$$\begin{cases}2 x_{1} + x_{2} - x_{3} + x_{4} = 1\\3 x_{2} - x_{4} = 1\\x_{3} - x_{4} = 0\end{cases}$$
Rapid solution
Gaussian elimination
Given the system of equations
$$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$
$$3 x_{2} - x_{4} = 1$$
$$x_{3} - x_{4} = 0$$
We give the system of equations to the canonical form
$$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$
$$3 x_{2} - x_{4} = 1$$
$$x_{3} - x_{4} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 1 & -1 & 1 & 1\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & -1 & 1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}1\\3\\0\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & 0 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-1\\0\\1\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}0\\3\\0\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & 0 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
In 3 -th column
$$\left[\begin{matrix}0\\0\\1\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
Everything is almost ready - it remains only to find the unknowns by solving elementary equations:
$$2 x_{1} + \frac{x_{4}}{3} - \frac{2}{3} = 0$$
$$3 x_{2} - x_{4} - 1 = 0$$
$$x_{3} - x_{4} = 0$$
We get the answer:
$$x_{1} = - \frac{x_{4}}{6} + \frac{1}{3}$$
$$x_{2} = \frac{x_{4}}{3} + \frac{1}{3}$$
$$x_{3} = x_{4}$$
where x4 - the free variables
$$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$
$$3 x_{2} - x_{4} = 1$$
$$x_{3} - x_{4} = 0$$
We give the system of equations to the canonical form
$$2 x_{1} + x_{2} - x_{3} + x_{4} = 1$$
$$3 x_{2} - x_{4} = 1$$
$$x_{3} - x_{4} = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 1 & -1 & 1 & 1\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & -1 & 1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}1\\3\\0\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & 0 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & -1 & \frac{4}{3} & \frac{2}{3}\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}-1\\0\\1\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\\0 & 3 & 0 & -1 & 1\\0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\0\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
In 2 -th column
$$\left[\begin{matrix}0\\3\\0\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 3 & 0 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
In 3 -th column
$$\left[\begin{matrix}0\\0\\1\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & 1 & -1 & 0\end{matrix}\right]$$
,
and subtract it from other lines:
Everything is almost ready - it remains only to find the unknowns by solving elementary equations:
$$2 x_{1} + \frac{x_{4}}{3} - \frac{2}{3} = 0$$
$$3 x_{2} - x_{4} - 1 = 0$$
$$x_{3} - x_{4} = 0$$
We get the answer:
$$x_{1} = - \frac{x_{4}}{6} + \frac{1}{3}$$
$$x_{2} = \frac{x_{4}}{3} + \frac{1}{3}$$
$$x_{3} = x_{4}$$
where x4 - the free variables