Question 87891
[7/2x]+[1/2x]=[(3x+3)/2]+[5/2x]

 [7/2x]+[1/2x]=[x(3x+3)/2x]+[5/2x]

 [7+1]/2x = [3x^2+3x+5]/2x

 3x^2+3x+5 = 8

 3x^2+3x-3 = 0

x^2+x-1 = 0

x = [-1+-sqrt(1-4*1*-1)]/2

x = [-1 +- sqrt5]/2
x= [-1+srt5]/2 or x= [-1-sqrt5]/2

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Cheers,
Stan H.