Mister Exam
Other calculators
- Inequality Systems step by step
- Complex numbers step by step
- Differential equations Step by Step
- The system of differential equations in steps
-
How to use it?
- 2x+y=16; 2x-y=24
- у/х-ху=-10; 5у/х-2ху=13
- 3^(x+2*y)=81; 0,1^(x)*10^(3*y)=10
- 4х+5y+4z=3; -7x+2y-z=2; 3x+y+z=-1
- 2x+y
- Plot:
-
2x+y
- Canonical form:
- 2x+y
-
Identical expressions
- 2x+y= sixteen ; 2x-y= twenty-four
- 2x plus y equally 16; 2x minus y equally 24
- 2x plus y equally sixteen ; 2x minus y equally twenty minus four
-
Similar expressions
- 2x+y=16; 2x+y=24
- 2x-y=16; 2x-y=24
2x+y=16; 2x-y=24
The solution
Detail solution
Given the system of equations
$$2 x + y = 16$$
$$2 x - y = 24$$
Let's express from equation 1 x
$$2 x + y = 16$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 16 - y$$
$$2 x = 16 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{16 - y}{2}$$
$$x = 8 - \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$2 x - y = 24$$
We get:
$$- y + 2 \left(8 - \frac{y}{2}\right) = 24$$
$$16 - 2 y = 24$$
We move the free summand 16 from the left part to the right part performing the sign change
$$- 2 y = -16 + 24$$
$$- 2 y = 8$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 2 y}{-2} = \frac{8}{-2}$$
$$y = -4$$
Because
$$x = 8 - \frac{y}{2}$$
then
$$x = 8 - -2$$
$$x = 10$$
The answer:
$$x = 10$$
$$y = -4$$
$$2 x + y = 16$$
$$2 x - y = 24$$
Let's express from equation 1 x
$$2 x + y = 16$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$2 x = 16 - y$$
$$2 x = 16 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{2 x}{2} = \frac{16 - y}{2}$$
$$x = 8 - \frac{y}{2}$$
Let's try the obtained element x to 2-th equation
$$2 x - y = 24$$
We get:
$$- y + 2 \left(8 - \frac{y}{2}\right) = 24$$
$$16 - 2 y = 24$$
We move the free summand 16 from the left part to the right part performing the sign change
$$- 2 y = -16 + 24$$
$$- 2 y = 8$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 2 y}{-2} = \frac{8}{-2}$$
$$y = -4$$
Because
$$x = 8 - \frac{y}{2}$$
then
$$x = 8 - -2$$
$$x = 10$$
The answer:
$$x = 10$$
$$y = -4$$
Rapid solution
$$x_{1} = 10$$
=
$$10$$
=
$$y_{1} = -4$$
=
$$-4$$
=
=
$$10$$
=
10
$$y_{1} = -4$$
=
$$-4$$
=
-4
Cramer's rule
$$2 x + y = 16$$
$$2 x - y = 24$$
We give the system of equations to the canonical form
$$2 x + y = 16$$
$$2 x - y = 24$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + x_{2}\\2 x_{1} - x_{2}\end{matrix}\right] = \left[\begin{matrix}16\\24\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 1\\2 & -1\end{matrix}\right] \right)} = -4$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}16 & 1\\24 & -1\end{matrix}\right] \right)}}{4} = 10$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 16\\2 & 24\end{matrix}\right] \right)}}{4} = -4$$
$$2 x - y = 24$$
We give the system of equations to the canonical form
$$2 x + y = 16$$
$$2 x - y = 24$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 x_{1} + x_{2}\\2 x_{1} - x_{2}\end{matrix}\right] = \left[\begin{matrix}16\\24\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}2 & 1\\2 & -1\end{matrix}\right] \right)} = -4$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}16 & 1\\24 & -1\end{matrix}\right] \right)}}{4} = 10$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}2 & 16\\2 & 24\end{matrix}\right] \right)}}{4} = -4$$
Gaussian elimination
Given the system of equations
$$2 x + y = 16$$
$$2 x - y = 24$$
We give the system of equations to the canonical form
$$2 x + y = 16$$
$$2 x - y = 24$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 1 & 16\\2 & -1 & 24\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & 16\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & -1 - 1 & \left(-1\right) 16 + 24\end{matrix}\right] = \left[\begin{matrix}0 & -2 & 8\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 1 & 16\\0 & -2 & 8\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\-2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -2 & 8\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-1\right) 0}{2} & 1 - - -1 & 16 - \frac{\left(-1\right) 8}{2}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 20\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 20\\0 & -2 & 8\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 20 = 0$$
$$- 2 x_{2} - 8 = 0$$
We get the answer:
$$x_{1} = 10$$
$$x_{2} = -4$$
$$2 x + y = 16$$
$$2 x - y = 24$$
We give the system of equations to the canonical form
$$2 x + y = 16$$
$$2 x - y = 24$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & 1 & 16\\2 & -1 & 24\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\2\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & 1 & 16\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & -1 - 1 & \left(-1\right) 16 + 24\end{matrix}\right] = \left[\begin{matrix}0 & -2 & 8\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 1 & 16\\0 & -2 & 8\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\-2\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -2 & 8\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-1\right) 0}{2} & 1 - - -1 & 16 - \frac{\left(-1\right) 8}{2}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 20\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 20\\0 & -2 & 8\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 20 = 0$$
$$- 2 x_{2} - 8 = 0$$
We get the answer:
$$x_{1} = 10$$
$$x_{2} = -4$$