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?
- 18y+zx=0; 8x+zy=0; xy-1.5=0
- x-1=1+0.5*y; x+y=1
- 17=0,27a-b; 75=2,19a+b
- y^2-180y+2xy=0; x^2-180x+2xy=0
- Graphing y =:
- x-1
- Canonical form:
- x-1
- Derivative of:
-
x-1
-
Identical expressions
- x- one = one + zero . five *y; x+y= one
- x minus 1 equally 1 plus 0.5 multiply by y; x plus y equally 1
- x minus one equally one plus zero . five multiply by y; x plus y equally one
- x-1=1+0.5y; x+y=1
-
Similar expressions
- x-1=1-0.5*y; x+y=1
- x-1=1+0.5*y; x-y=1
- x+1=1+0.5*y; x+y=1
x-1=1+0.5*y; x+y=1
The solution
You have entered
[src]
y
x - 1 = 1 + -
2
$$x - 1 = \frac{y}{2} + 1$$
x + y = 1
$$x + y = 1$$
x + y = 1
Detail solution
Given the system of equations
$$x - 1 = \frac{y}{2} + 1$$
$$x + y = 1$$
Let's express from equation 1 x
$$x - 1 = \frac{y}{2} + 1$$
We move the free summand -1 from the left part to the right part performing the sign change
$$x = \left(\frac{y}{2} + 1\right) + 1$$
$$x = \frac{y}{2} + 2$$
Let's try the obtained element x to 2-th equation
$$x + y = 1$$
We get:
$$y + \left(\frac{y}{2} + 2\right) = 1$$
$$\frac{3 y}{2} + 2 = 1$$
We move the free summand 2 from the left part to the right part performing the sign change
$$\frac{3 y}{2} = -2 + 1$$
$$\frac{3 y}{2} = -1$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{3}{2} y}{\frac{3}{2}} = - \frac{1}{\frac{3}{2}}$$
$$y = - \frac{2}{3}$$
Because
$$x = \frac{y}{2} + 2$$
then
$$x = \frac{-2}{2 \cdot 3} + 2$$
$$x = \frac{5}{3}$$
The answer:
$$x = \frac{5}{3}$$
$$y = - \frac{2}{3}$$
$$x - 1 = \frac{y}{2} + 1$$
$$x + y = 1$$
Let's express from equation 1 x
$$x - 1 = \frac{y}{2} + 1$$
We move the free summand -1 from the left part to the right part performing the sign change
$$x = \left(\frac{y}{2} + 1\right) + 1$$
$$x = \frac{y}{2} + 2$$
Let's try the obtained element x to 2-th equation
$$x + y = 1$$
We get:
$$y + \left(\frac{y}{2} + 2\right) = 1$$
$$\frac{3 y}{2} + 2 = 1$$
We move the free summand 2 from the left part to the right part performing the sign change
$$\frac{3 y}{2} = -2 + 1$$
$$\frac{3 y}{2} = -1$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{3}{2} y}{\frac{3}{2}} = - \frac{1}{\frac{3}{2}}$$
$$y = - \frac{2}{3}$$
Because
$$x = \frac{y}{2} + 2$$
then
$$x = \frac{-2}{2 \cdot 3} + 2$$
$$x = \frac{5}{3}$$
The answer:
$$x = \frac{5}{3}$$
$$y = - \frac{2}{3}$$
Rapid solution
$$x_{1} = \frac{5}{3}$$
=
$$\frac{5}{3}$$
=
$$y_{1} = - \frac{2}{3}$$
=
$$- \frac{2}{3}$$
=
=
$$\frac{5}{3}$$
=
1.66666666666667
$$y_{1} = - \frac{2}{3}$$
=
$$- \frac{2}{3}$$
=
-0.666666666666667
Gaussian elimination
Given the system of equations
$$x - 1 = \frac{y}{2} + 1$$
$$x + y = 1$$
We give the system of equations to the canonical form
$$x - \frac{y}{2} = 2$$
$$x + y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & - \frac{1}{2} & 2\\1 & 1 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & - \frac{1}{2} & 2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & 1 - - \frac{1}{2} & \left(-1\right) 2 + 1\end{matrix}\right] = \left[\begin{matrix}0 & \frac{3}{2} & -1\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & - \frac{1}{2} & 2\\0 & \frac{3}{2} & -1\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{2}\\\frac{3}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{3}{2} & -1\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{3} & - \frac{1}{2} - \frac{\left(-1\right) 3}{2 \cdot 3} & 2 - - \frac{-1}{3}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & \frac{5}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & \frac{5}{3}\\0 & \frac{3}{2} & -1\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - \frac{5}{3} = 0$$
$$\frac{3 x_{2}}{2} + 1 = 0$$
We get the answer:
$$x_{1} = \frac{5}{3}$$
$$x_{2} = - \frac{2}{3}$$
$$x - 1 = \frac{y}{2} + 1$$
$$x + y = 1$$
We give the system of equations to the canonical form
$$x - \frac{y}{2} = 2$$
$$x + y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & - \frac{1}{2} & 2\\1 & 1 & 1\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\1\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & - \frac{1}{2} & 2\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & 1 - - \frac{1}{2} & \left(-1\right) 2 + 1\end{matrix}\right] = \left[\begin{matrix}0 & \frac{3}{2} & -1\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & - \frac{1}{2} & 2\\0 & \frac{3}{2} & -1\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{2}\\\frac{3}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{3}{2} & -1\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-1\right) 0}{3} & - \frac{1}{2} - \frac{\left(-1\right) 3}{2 \cdot 3} & 2 - - \frac{-1}{3}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & \frac{5}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & \frac{5}{3}\\0 & \frac{3}{2} & -1\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - \frac{5}{3} = 0$$
$$\frac{3 x_{2}}{2} + 1 = 0$$
We get the answer:
$$x_{1} = \frac{5}{3}$$
$$x_{2} = - \frac{2}{3}$$
Cramer's rule
$$x - 1 = \frac{y}{2} + 1$$
$$x + y = 1$$
We give the system of equations to the canonical form
$$x - \frac{y}{2} = 2$$
$$x + y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - \frac{x_{2}}{2}\\x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}2\\1\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & - \frac{1}{2}\\1 & 1\end{matrix}\right] \right)} = \frac{3}{2}$$
, 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 )
$$x_{1} = \frac{2 \operatorname{det}{\left(\left[\begin{matrix}2 & - \frac{1}{2}\\1 & 1\end{matrix}\right] \right)}}{3} = \frac{5}{3}$$
$$x_{2} = \frac{2 \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\1 & 1\end{matrix}\right] \right)}}{3} = - \frac{2}{3}$$
$$x + y = 1$$
We give the system of equations to the canonical form
$$x - \frac{y}{2} = 2$$
$$x + y = 1$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}x_{1} - \frac{x_{2}}{2}\\x_{1} + x_{2}\end{matrix}\right] = \left[\begin{matrix}2\\1\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}1 & - \frac{1}{2}\\1 & 1\end{matrix}\right] \right)} = \frac{3}{2}$$
, 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 )
$$x_{1} = \frac{2 \operatorname{det}{\left(\left[\begin{matrix}2 & - \frac{1}{2}\\1 & 1\end{matrix}\right] \right)}}{3} = \frac{5}{3}$$
$$x_{2} = \frac{2 \operatorname{det}{\left(\left[\begin{matrix}1 & 2\\1 & 1\end{matrix}\right] \right)}}{3} = - \frac{2}{3}$$
Numerical answer
[src]
x1 = 1.666666666666667 y1 = -0.6666666666666667
x1 = 1.666666666666667 y1 = -0.6666666666666667