(7/y)/(4/y+5)

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Solution for (7/y)/(4/y+5) equation: 


D( y )

y = 0

4/y+5 = 0

y = 0

y = 0

4/y+5 = 0

4/y+5 = 0

4*y^-1 = -5 // : 4

y^-1 = -5/4

-1 < 0

1/(y^1) = -5/4 // * y^1

1 = -5/4*y^1 // : -5/4

-4/5 = y^1

y = -4/5

y in (-oo:-4/5) U (-4/5:0) U (0:+oo)

(7/y)/(4/y+5) = 0

(7*y^-1)/(4*y^-1+5) = 0

7*y^-1 = 0 // : 7

y^-1 = 0

y naleu017Cy do O

y belongs to the empty set

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