(5/y+2)+(2/y-3)=(7/y+6)

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Solution for (5/y+2)+(2/y-3)=(7/y+6) equation: 


D( y )

y = 0

y = 0

y = 0

y in (-oo:0) U (0:+oo)

5/y+2/y-3+2 = 7/y+6 // - 7/y+6

5/y+2/y-(7/y)-6-3+2 = 0

5/y+2/y-7*y^-1-6-3+2 = 0

-7 = 0

y belongs to the empty set

y belongs to the empty set

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