1/(2x-y)=-1 equation
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The solution
Detail solution
Given the equation:
$$\frac{1}{2 x - y} = -1$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
a1 = 1
b1 = -y + 2*x
a2 = 1
b2 = -1
so we get the equation
$$-1 = 2 x - y$$
$$-1 = 2 x - y$$
Looking for similar summands in the right part:
-1 = -y + 2*x
Move free summands (without x)
from left part to right part, we given:
$$0 = 2 x - y + 1$$
Divide both parts of the equation by 0
x = 1 - y + 2*x / (0)
We get the answer: x = -1/2 + y/2
1 re(y) I*im(y)
x1 = - - + ----- + -------
2 2 2
$$x_{1} = \frac{\operatorname{re}{\left(y\right)}}{2} + \frac{i \operatorname{im}{\left(y\right)}}{2} - \frac{1}{2}$$
x1 = re(y)/2 + i*im(y)/2 - 1/2
Sum and product of roots
[src]
1 re(y) I*im(y)
- - + ----- + -------
2 2 2
$$\frac{\operatorname{re}{\left(y\right)}}{2} + \frac{i \operatorname{im}{\left(y\right)}}{2} - \frac{1}{2}$$
1 re(y) I*im(y)
- - + ----- + -------
2 2 2
$$\frac{\operatorname{re}{\left(y\right)}}{2} + \frac{i \operatorname{im}{\left(y\right)}}{2} - \frac{1}{2}$$
1 re(y) I*im(y)
- - + ----- + -------
2 2 2
$$\frac{\operatorname{re}{\left(y\right)}}{2} + \frac{i \operatorname{im}{\left(y\right)}}{2} - \frac{1}{2}$$
1 re(y) I*im(y)
- - + ----- + -------
2 2 2
$$\frac{\operatorname{re}{\left(y\right)}}{2} + \frac{i \operatorname{im}{\left(y\right)}}{2} - \frac{1}{2}$$
-1/2 + re(y)/2 + i*im(y)/2