Mister Exam
X+2/6-y-3/4=1; X-2/4-y-4/2=1
The solution
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x + 1/3 - y - 3/4 = 1
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
x - 1/2 - y - 2 = 1
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
-y + x - 1/2 - 2 = 1
Detail solution
Given the system of equations
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
Let's express from equation 1 x
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x - \frac{5}{12} = y + 1$$
$$x - \frac{5}{12} = y + 1$$
We move the free summand -5/12 from the left part to the right part performing the sign change
$$x = \left(y + 1\right) + \frac{5}{12}$$
$$x = y + \frac{17}{12}$$
Let's try the obtained element x to 2-th equation
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
We get:
$$\left(- y + \left(\left(y + \frac{17}{12}\right) - \frac{1}{2}\right)\right) - 2 = 1$$
so
This system of equations has no solutions
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
Let's express from equation 1 x
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x - \frac{5}{12} = y + 1$$
$$x - \frac{5}{12} = y + 1$$
We move the free summand -5/12 from the left part to the right part performing the sign change
$$x = \left(y + 1\right) + \frac{5}{12}$$
$$x = y + \frac{17}{12}$$
Let's try the obtained element x to 2-th equation
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
We get:
$$\left(- y + \left(\left(y + \frac{17}{12}\right) - \frac{1}{2}\right)\right) - 2 = 1$$
so
This system of equations has no solutions
Gaussian elimination
Given the system of equations
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
We give the system of equations to the canonical form
$$x - y = \frac{17}{12}$$
$$x - y = \frac{7}{2}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\\1 & -1 & \frac{7}{2}\end{matrix}\right]$$
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
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & -1 - -1 & \frac{\left(-1\right) 17}{12} + \frac{7}{2}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{25}{12}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\\0 & 0 & \frac{25}{12}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$x_{1} - x_{2} - \frac{17}{12} = 0$$
$$0 - 25/12 = 0$$
We get the answer:
This system of equations has no solutions
$$\left(- y + \left(x + \frac{1}{3}\right)\right) - \frac{3}{4} = 1$$
$$\left(- y + \left(x - \frac{1}{2}\right)\right) - 2 = 1$$
We give the system of equations to the canonical form
$$x - y = \frac{17}{12}$$
$$x - y = \frac{7}{2}$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\\1 & -1 & \frac{7}{2}\end{matrix}\right]$$
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
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-1 + 1 & -1 - -1 & \frac{\left(-1\right) 17}{12} + \frac{7}{2}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{25}{12}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\\0 & 0 & \frac{25}{12}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & -1 & \frac{17}{12}\end{matrix}\right]$$
,
and subtract it from other lines:
We prepare elementary equations using a solved matrix and see that this system of equations has no decisions
$$x_{1} - x_{2} - \frac{17}{12} = 0$$
$$0 - 25/12 = 0$$
We get the answer:
This system of equations has no solutions