(1)/(8)(3d-2)=(1)/(4)(d+5)

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Solution for (1)/(8)(3d-2)=(1)/(4)(d+5) equation:



(1)/(8)(3d-2)=(1)/(4)(d+5)
We move all terms to the left:
(1)/(8)(3d-2)-((1)/(4)(d+5))=0
Domain of the equation: 8(3d-2)!=0
d∈R
Domain of the equation: 4(d+5))!=0
d∈R
We calculate fractions
(4dd/(8(3d-2)*4(d+5)))+(-8d3/(8(3d-2)*4(d+5)))=0
We calculate terms in parentheses: +(4dd/(8(3d-2)*4(d+5))), so:
4dd/(8(3d-2)*4(d+5))
We multiply all the terms by the denominator
4dd
Back to the equation:
+(4dd)
We calculate terms in parentheses: +(-8d3/(8(3d-2)*4(d+5))), so:
-8d3/(8(3d-2)*4(d+5))
We multiply all the terms by the denominator
-8d3
We add all the numbers together, and all the variables
-8d^3
We do not support edpression: d^3

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