3/(n+1)=4/(n+4)

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Solution for 3/(n+1)=4/(n+4) equation:


D( n )

n+4 = 0

n+1 = 0

n+4 = 0

n+4 = 0

n+4 = 0 // - 4

n = -4

n+1 = 0

n+1 = 0

n+1 = 0 // - 1

n = -1

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

3/(n+1) = 4/(n+4) // - 4/(n+4)

3/(n+1)-(4/(n+4)) = 0

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

3/(n+1)-4/(n+4) = 0

(3*(n+4))/((n+1)*(n+4))+(-4*(n+1))/((n+1)*(n+4)) = 0

3*(n+4)-4*(n+1) = 0

8-n = 0

(8-n)/((n+1)*(n+4)) = 0

(8-n)/((n+1)*(n+4)) = 0 // * (n+1)*(n+4)

8-n = 0

8-n = 0 // - 8

-n = -8 // * -1

n = 8

n = 8

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