2(1/6)=1(2/3)m+2

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Solution for 2(1/6)=1(2/3)m+2 equation:



2(1/6)=1(2/3)m+2
We move all terms to the left:
2(1/6)-(1(2/3)m+2)=0
Domain of the equation: 3)m+2)!=0
m!=0/1
m!=0
m∈R
We add all the numbers together, and all the variables
-(1(+2/3)m+2)+2(+1/6)=0
We multiply parentheses
-(1(+2/3)m+2)+1/6*2=0
We calculate fractions
()/3m+2)*6*2)+3m/3m+2)*6*2)=0
We calculate fractions
(()*3m)/9m^2+2*6*2)+3m*3m)/9m^2=0
We multiply all the terms by the denominator
(()*3m)+2*6*2)+3m*3m)=0
We calculate terms in parentheses: +(()*3m), so:
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Wy multiply elements
72m^2*2+(()*3m)=0
We calculate terms in parentheses: +(()*3m), so:
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Wy multiply elements
144m^2+(()*3m)=0
We calculate terms in parentheses: +(()*3m), so:
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