1/2(10y+2)+5y=41

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Solution for 1/2(10y+2)+5y=41 equation: 



1/2(10y+2)+5y=41

We move all terms to the left:

1/2(10y+2)+5y-(41)=0



Domain of the equation: 2(10y+2)!=0

y∈R



We add all the numbers together, and all the variables

5y+1/2(10y+2)-41=0

We multiply all the terms by the denominator


5y*2(10y+2)-41*2(10y+2)+1=0

Wy multiply elements

10y^2(1-82y(1+1=0

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