5(x-4y+2c)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the linear equation:
5*(x-4*y+2*c) = 0
Expand brackets in the left part
5*x-5*4*y+5*2*c = 0
Looking for similar summands in the left part:
-20*y + 5*x + 10*c = 0
Move the summands with the other variables
from left part to right part, we given:
$$10 c + 5 x = 20 y$$
Divide both parts of the equation by (5*x + 10*c)/x
x = 20*y / ((5*x + 10*c)/x)
We get the answer: x = -2*c + 4*y
x1 = -2*re(c) + 4*re(y) + I*(-2*im(c) + 4*im(y))
$$x_{1} = i \left(- 2 \operatorname{im}{\left(c\right)} + 4 \operatorname{im}{\left(y\right)}\right) - 2 \operatorname{re}{\left(c\right)} + 4 \operatorname{re}{\left(y\right)}$$
x1 = i*(-2*im(c) + 4*im(y)) - 2*re(c) + 4*re(y)
Sum and product of roots
[src]
-2*re(c) + 4*re(y) + I*(-2*im(c) + 4*im(y))
$$i \left(- 2 \operatorname{im}{\left(c\right)} + 4 \operatorname{im}{\left(y\right)}\right) - 2 \operatorname{re}{\left(c\right)} + 4 \operatorname{re}{\left(y\right)}$$
-2*re(c) + 4*re(y) + I*(-2*im(c) + 4*im(y))
$$i \left(- 2 \operatorname{im}{\left(c\right)} + 4 \operatorname{im}{\left(y\right)}\right) - 2 \operatorname{re}{\left(c\right)} + 4 \operatorname{re}{\left(y\right)}$$
-2*re(c) + 4*re(y) + I*(-2*im(c) + 4*im(y))
$$i \left(- 2 \operatorname{im}{\left(c\right)} + 4 \operatorname{im}{\left(y\right)}\right) - 2 \operatorname{re}{\left(c\right)} + 4 \operatorname{re}{\left(y\right)}$$
-2*re(c) + 4*re(y) + 2*I*(-im(c) + 2*im(y))
$$2 i \left(- \operatorname{im}{\left(c\right)} + 2 \operatorname{im}{\left(y\right)}\right) - 2 \operatorname{re}{\left(c\right)} + 4 \operatorname{re}{\left(y\right)}$$
-2*re(c) + 4*re(y) + 2*i*(-im(c) + 2*im(y))