(7x-3)/2=(3-9x)/6+5

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Solution for (7x-3)/2=(3-9x)/6+5 equation: 



(7x-3)/2=(3-9x)/6+5

We move all terms to the left:

(7x-3)/2-((3-9x)/6+5)=0

We add all the numbers together, and all the variables

(7x-3)/2-((-9x+3)/6+5)=0

We calculate fractions


(42x-18)/()+(-((-9x+3)*2)/()=0



We calculate terms in parentheses: +(-((-9x+3)*2)/(), so:

-((-9x+3)*2)/(

We multiply all the terms by the denominator


-((-9x+3)*2)



We calculate terms in parentheses: -((-9x+3)*2), so:

(-9x+3)*2

We multiply parentheses

-18x+6

Back to the equation:

-(-18x+6)



We get rid of parentheses

18x-6

Back to the equation:

+(18x-6)



We get rid of parentheses

(42x-18)/()+18x-6=0

We multiply all the terms by the denominator


(42x-18)+18x*()-6*()=0

We add all the numbers together, and all the variables

(42x-18)+18x*()=0

We get rid of parentheses

42x+18x*()-18=0

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

42x+18x*()=18

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