(5)/(n+2)-3=(1)/(n+2)

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Solution for (5)/(n+2)-3=(1)/(n+2) equation: 


D( n )

n+2 = 0

n+2 = 0

n+2 = 0

n+2 = 0 // - 2

n = -2

n in (-oo:-2) U (-2:+oo)

5/(n+2)-3 = 1/(n+2) // - 1/(n+2)

5/(n+2)-(1/(n+2))-3 = 0

5/(n+2)-(n+2)^-1-3 = 0

5/(n+2)-1/(n+2)-3 = 0

5/(n+2)-1/(n+2)+(-3*(n+2))/(n+2) = 0

5-3*(n+2)-1 = 0

4-3*n-6 = 0

-3*n-2 = 0

(-3*n-2)/(n+2) = 0

(-3*n-2)/(n+2) = 0 // * n+2

-3*n-2 = 0

-3*n-2 = 0 // + 2

-3*n = 2 // : -3

n = 2/(-3)

n = -2/3

n = -2/3

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