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

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


D( n )

n-1 = 0

n+3 = 0

n-1 = 0

n-1 = 0

n-1 = 0 // + 1

n = 1

n+3 = 0

n+3 = 0

n+3 = 0 // - 3

n = -3

n in (-oo:-3) U (-3:1) U (1:+oo)

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

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

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

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

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

3*n^2-5*n^2+24*n-10*n+15+21 = 0

14*n-2*n^2+36 = 0

14*n-2*n^2+36 = 0

2*(7*n-n^2+18) = 0

7*n-n^2+18 = 0

DELTA = 7^2-(-1*4*18)

DELTA = 121

DELTA > 0

n = (121^(1/2)-7)/(-1*2) or n = (-121^(1/2)-7)/(-1*2)

n = -2 or n = 9

2*(n+2)*(n-9) = 0

(2*(n+2)*(n-9))/(3*(n-1)*(n+3)) = 0

(2*(n+2)*(n-9))/(3*(n-1)*(n+3)) = 0 // * 3*(n-1)*(n+3)

2*(n+2)*(n-9) = 0

( n+2 )

n+2 = 0 // - 2

n = -2

( n-9 )

n-9 = 0 // + 9

n = 9

n in { -2, 9 }

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