(3x-7)/5=(2x-2)/4+0,2

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Solution for (3x-7)/5=(2x-2)/4+0,2 equation: 



(3x-7)/5=(2x-2)/4+0.2

We move all terms to the left:

(3x-7)/5-((2x-2)/4+0.2)=0

We calculate fractions


(12x-28)/()+(-((2x-2)*5)/()=0



We calculate terms in parentheses: +(-((2x-2)*5)/(), so:

-((2x-2)*5)/(

We multiply all the terms by the denominator


-((2x-2)*5)



We calculate terms in parentheses: -((2x-2)*5), so:

(2x-2)*5

We multiply parentheses

10x-10

Back to the equation:

-(10x-10)



We get rid of parentheses

-10x+10

Back to the equation:

+(-10x+10)



We get rid of parentheses

(12x-28)/()-10x+10=0

We multiply all the terms by the denominator


(12x-28)-10x*()+10*()=0

We add all the numbers together, and all the variables

(12x-28)-10x*()=0

We get rid of parentheses

12x-10x*()-28=0

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

12x-10x*()=28

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