4+((3)/(u-1))=((7)/(u+4))

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Solution for 4+((3)/(u-1))=((7)/(u+4)) equation: 


D( u )

u+4 = 0

u-1 = 0

u+4 = 0

u+4 = 0

u+4 = 0 // - 4

u = -4

u-1 = 0

u-1 = 0

u-1 = 0 // + 1

u = 1

u in (-oo:-4) U (-4:1) U (1:+oo)

3/(u-1)+4 = 7/(u+4) // - 7/(u+4)

3/(u-1)-(7/(u+4))+4 = 0

3/(u-1)-7*(u+4)^-1+4 = 0

3/(u-1)-7/(u+4)+4 = 0

(3*(u+4))/((u-1)*(u+4))+(-7*(u-1))/((u-1)*(u+4))+(4*(u-1)*(u+4))/((u-1)*(u+4)) = 0

3*(u+4)-7*(u-1)+4*(u-1)*(u+4) = 0

4*u^2-4*u+12*u-16+19 = 0

4*u^2+8*u+3 = 0

4*u^2+8*u+3 = 0

4*u^2+8*u+3 = 0

DELTA = 8^2-(3*4*4)

DELTA = 16

DELTA > 0

u = (16^(1/2)-8)/(2*4) or u = (-16^(1/2)-8)/(2*4)

u = -1/2 or u = -3/2

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

((u+3/2)*(u+1/2))/((u-1)*(u+4)) = 0

((u+3/2)*(u+1/2))/((u-1)*(u+4)) = 0 // * (u-1)*(u+4)

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

( u+1/2 )

u+1/2 = 0 // - 1/2

u = -1/2

( u+3/2 )

u+3/2 = 0 // - 3/2

u = -3/2

u in { -1/2, -3/2 }

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