(u+2)/(3u-1)+1=(6-u)/(u+1)

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Solution for (u+2)/(3u-1)+1=(6-u)/(u+1) equation: 


D( u )

u+1 = 0

3*u-1 = 0

u+1 = 0

u+1 = 0

u+1 = 0 // - 1

u = -1

3*u-1 = 0

3*u-1 = 0

3*u-1 = 0 // + 1

3*u = 1 // : 3

u = 1/3

u in (-oo:-1) U (-1:1/3) U (1/3:+oo)

(u+2)/(3*u-1)+1 = (6-u)/(u+1) // - (6-u)/(u+1)

(u+2)/(3*u-1)-((6-u)/(u+1))+1 = 0

(u+2)/(3*u-1)+(-1*(6-u))/(u+1)+1 = 0

((u+2)*(u+1))/((3*u-1)*(u+1))+(-1*(6-u)*(3*u-1))/((3*u-1)*(u+1))+(1*(3*u-1)*(u+1))/((3*u-1)*(u+1)) = 0

(u+2)*(u+1)-1*(6-u)*(3*u-1)+1*(3*u-1)*(u+1) = 0

4*u^2+3*u^2-16*u+2*u-1+8 = 0

7*u^2-14*u+7 = 0

7*u^2-14*u+7 = 0

7*(u^2-2*u+1) = 0

u^2-2*u+1 = 0

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

DELTA = 0

u = 2/(1*2)

u = 1 or u = 1

7*(u-1)^2 = 0

(7*(u-1)^2)/((3*u-1)*(u+1)) = 0

(7*(u-1)^2)/((3*u-1)*(u+1)) = 0 // * (3*u-1)*(u+1)

7*(u-1)^2 = 0

u-1 = 0 // + 1

u = 1

u in { 1, 1 }

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