2x+(1/5)=(x-10)/3

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Solution for 2x+(1/5)=(x-10)/3 equation: 



2x+(1/5)=(x-10)/3

We move all terms to the left:

2x+(1/5)-((x-10)/3)=0

We add all the numbers together, and all the variables

2x-((x-10)/3)+(+1/5)=0

We get rid of parentheses

2x-((x-10)/3)+1/5=0

We calculate fractions


2x+(-((x-10)*5)/()+()/()=0



We calculate terms in parentheses: +(-((x-10)*5)/()+()/(), so:

-((x-10)*5)/()+()/(

We add all the numbers together, and all the variables

-((x-10)*5)/()+1

We multiply all the terms by the denominator


-((x-10)*5)+1*()



We calculate terms in parentheses: -((x-10)*5), so:

(x-10)*5

We multiply parentheses

5x-50

Back to the equation:

-(5x-50)



We add all the numbers together, and all the variables

-(5x-50)

We get rid of parentheses

-5x+50

Back to the equation:

+(-5x+50)



We get rid of parentheses

2x-5x+50=0

We add all the numbers together, and all the variables

-3x+50=0

We move all terms containing x to the left, all other terms to the right

-3x=-50

x=-50/-3

x=16+2/3

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