1/5(4x+2)=67

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Solution for 1/5(4x+2)=67 equation: 



1/5(4x+2)=67

We move all terms to the left:

1/5(4x+2)-(67)=0



Domain of the equation: 5(4x+2)!=0

x∈R



We multiply all the terms by the denominator


-67*5(4x+2)+1=0

Wy multiply elements

-335x(4+1=0

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