Mister Exam
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- Inequality Systems step by step
- Complex numbers step by step
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- The system of differential equations in steps
-
How to use it?
- x+y=-2; 6*х+5у=36
- 4-x=y+5; y=14+4x
- 2/у-1+3/х+1=5/2; 1/х-1=-3/у
- х^2+ху-3у=-1; 4х-у=3
- Graphing y =:
- 4-x
- Derivative of:
-
4-x
- Limit of the function:
-
4-x
-
Identical expressions
- four -x=y+ five ; y= fourteen +4x
- 4 minus x equally y plus 5; y equally 14 plus 4x
- four minus x equally y plus five ; y equally fourteen plus 4x
-
Similar expressions
- log3(5x+4y)=log3(y+5); y-2x=2
- 4+x=y+5; y=14+4x
- 4-x=y+5; y=14-4x
- 4x-y-24=2(5x-2); 3y-2=4-(x-y)
- 4-x=y-5; y=14+4x
4-x=y+5; y=14+4x
The solution
Detail solution
Given the system of equations
$$4 - x = y + 5$$
$$y = 4 x + 14$$
Let's express from equation 1 x
$$4 - x = y + 5$$
We move the free summand 4 from the left part to the right part performing the sign change
$$- x = \left(y + 5\right) - 4$$
$$- x = y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\left(-1\right) x}{-1} = \frac{y + 1}{-1}$$
$$x = - y - 1$$
Let's try the obtained element x to 2-th equation
$$y = 4 x + 14$$
We get:
$$y = 4 \left(- y - 1\right) + 14$$
$$y = 10 - 4 y$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$y + 4 y = 10$$
$$5 y = 10$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{5 y}{5} = \frac{10}{5}$$
$$y = 2$$
Because
$$x = - y - 1$$
then
$$x = - 2 - 1$$
$$x = -3$$
The answer:
$$x = -3$$
$$y = 2$$
$$4 - x = y + 5$$
$$y = 4 x + 14$$
Let's express from equation 1 x
$$4 - x = y + 5$$
We move the free summand 4 from the left part to the right part performing the sign change
$$- x = \left(y + 5\right) - 4$$
$$- x = y + 1$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\left(-1\right) x}{-1} = \frac{y + 1}{-1}$$
$$x = - y - 1$$
Let's try the obtained element x to 2-th equation
$$y = 4 x + 14$$
We get:
$$y = 4 \left(- y - 1\right) + 14$$
$$y = 10 - 4 y$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$y + 4 y = 10$$
$$5 y = 10$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{5 y}{5} = \frac{10}{5}$$
$$y = 2$$
Because
$$x = - y - 1$$
then
$$x = - 2 - 1$$
$$x = -3$$
The answer:
$$x = -3$$
$$y = 2$$
Rapid solution
$$x_{1} = -3$$
=
$$-3$$
=
$$y_{1} = 2$$
=
$$2$$
=
=
$$-3$$
=
-3
$$y_{1} = 2$$
=
$$2$$
=
2
Gaussian elimination
Given the system of equations
$$4 - x = y + 5$$
$$y = 4 x + 14$$
We give the system of equations to the canonical form
$$- x - y = 1$$
$$- 4 x + y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-1 & -1 & 1\\-4 & 1 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-1\\-4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-1 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-4 - - 4 & 1 - - 4 & \left(-1\right) 4 + 14\end{matrix}\right] = \left[\begin{matrix}0 & 5 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-1 & -1 & 1\\0 & 5 & 10\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 5 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 - \frac{\left(-1\right) 0}{5} & -1 - \frac{\left(-1\right) 5}{5} & 1 - \frac{\left(-1\right) 10}{5}\end{matrix}\right] = \left[\begin{matrix}-1 & 0 & 3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-1 & 0 & 3\\0 & 5 & 10\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- x_{1} - 3 = 0$$
$$5 x_{2} - 10 = 0$$
We get the answer:
$$x_{1} = -3$$
$$x_{2} = 2$$
$$4 - x = y + 5$$
$$y = 4 x + 14$$
We give the system of equations to the canonical form
$$- x - y = 1$$
$$- 4 x + y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-1 & -1 & 1\\-4 & 1 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-1\\-4\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-1 & -1 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-4 - - 4 & 1 - - 4 & \left(-1\right) 4 + 14\end{matrix}\right] = \left[\begin{matrix}0 & 5 & 10\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-1 & -1 & 1\\0 & 5 & 10\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-1\\5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 5 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 - \frac{\left(-1\right) 0}{5} & -1 - \frac{\left(-1\right) 5}{5} & 1 - \frac{\left(-1\right) 10}{5}\end{matrix}\right] = \left[\begin{matrix}-1 & 0 & 3\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-1 & 0 & 3\\0 & 5 & 10\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- x_{1} - 3 = 0$$
$$5 x_{2} - 10 = 0$$
We get the answer:
$$x_{1} = -3$$
$$x_{2} = 2$$
Cramer's rule
$$4 - x = y + 5$$
$$y = 4 x + 14$$
We give the system of equations to the canonical form
$$- x - y = 1$$
$$- 4 x + y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- x_{1} - x_{2}\\- 4 x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\14\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-1 & -1\\-4 & 1\end{matrix}\right] \right)} = -5$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -1\\14 & 1\end{matrix}\right] \right)}}{5} = -3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-1 & 1\\-4 & 14\end{matrix}\right] \right)}}{5} = 2$$
$$y = 4 x + 14$$
We give the system of equations to the canonical form
$$- x - y = 1$$
$$- 4 x + y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- x_{1} - x_{2}\\- 4 x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}1\\14\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-1 & -1\\-4 & 1\end{matrix}\right] \right)} = -5$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}1 & -1\\14 & 1\end{matrix}\right] \right)}}{5} = -3$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-1 & 1\\-4 & 14\end{matrix}\right] \right)}}{5} = 2$$