2/5x=1(1/6)

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Solution for 2/5x=1(1/6) equation:



2/5x=1(1/6)
We move all terms to the left:
2/5x-(1(1/6))=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
We add all the numbers together, and all the variables
2/5x-(1(+1/6))=0
We calculate fractions
()/30x^2+(-(1(+1*5x)/30x^2=0
We multiply all the terms by the denominator
(-(1(+1*5x)+()=0
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
-(1(+1*5x)+(
We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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We calculate terms in parentheses: +(-(1(+1*5x)+(), so:
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