2(7+2u)=1/8(32+32u)

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Solution for 2(7+2u)=1/8(32+32u) equation: 



2(7+2u)=1/8(32+32u)

We move all terms to the left:

2(7+2u)-(1/8(32+32u))=0



Domain of the equation: 8(32+32u))!=0

u∈R



We add all the numbers together, and all the variables

2(2u+7)-(1/8(32u+32))=0

We multiply parentheses

4u-(1/8(32u+32))+14=0

We multiply all the terms by the denominator


4u*8(32u+32))-(1+14*8(32u+32))=0

Wy multiply elements

32u^2(3+112u(3=0

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