1/5(2x+12)+4x=33

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Solution for 1/5(2x+12)+4x=33 equation: 



1/5(2x+12)+4x=33

We move all terms to the left:

1/5(2x+12)+4x-(33)=0



Domain of the equation: 5(2x+12)!=0

x∈R



We add all the numbers together, and all the variables

4x+1/5(2x+12)-33=0

We multiply all the terms by the denominator


4x*5(2x+12)-33*5(2x+12)+1=0

Wy multiply elements

20x^2(2-165x(2+1=0

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