3/(z-1)-1/5=4/(z+1)

Solution for 3/(z-1)-1/5=4/(z+1) equation:

D( z )

z-1 = 0

z+1 = 0

z-1 = 0

z-1 = 0

z-1 = 0 // + 1

z = 1

z+1 = 0

z+1 = 0

z+1 = 0 // - 1

z = -1

z in (-oo:-1) U (-1:1) U (1:+oo)

3/(z-1)-(1/5) = 4/(z+1) // - 4/(z+1)

3/(z-1)-(4/(z+1))-(1/5) = 0

3/(z-1)-4*(z+1)^-1-1/5 = 0

3/(z-1)-4/(z+1)-1/5 = 0

(3*5*(z+1))/(5*(z-1)*(z+1))+(-4*5*(z-1))/(5*(z-1)*(z+1))+(-1*(z-1)*(z+1))/(5*(z-1)*(z+1)) = 0

3*5*(z+1)-4*5*(z-1)-1*(z-1)*(z+1) = 0

1-z^2-5*z+35 = 0

36-z^2-5*z = 0

36-z^2-5*z = 0

36-z^2-5*z = 0

DELTA = (-5)^2-(-1*4*36)

DELTA = 169

DELTA > 0

z = (169^(1/2)+5)/(-1*2) or z = (5-169^(1/2))/(-1*2)

z = -9 or z = 4

(z+9)*(z-4) = 0

((z+9)*(z-4))/(5*(z-1)*(z+1)) = 0

((z+9)*(z-4))/(5*(z-1)*(z+1)) = 0 // * 5*(z-1)*(z+1)

(z+9)*(z-4) = 0

( z+9 )

z+9 = 0 // - 9

z = -9

( z-4 )

z-4 = 0 // + 4

z = 4

z in { -9, 4 }