(4(3x-5)+12)/8=(3x-12x)/6

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Solution for (4(3x-5)+12)/8=(3x-12x)/6 equation: 



(4(3x-5)+12)/8=(3x-12x)/6

We move all terms to the left:

(4(3x-5)+12)/8-((3x-12x)/6)=0

We add all the numbers together, and all the variables

(4(3x-5)+12)/8-((-9x)/6)=0

We calculate fractions


(12x-20)/()+(-((-9x)*8)/()=0



We calculate terms in parentheses: +(-((-9x)*8)/(), so:

-((-9x)*8)/(

We multiply all the terms by the denominator


-((-9x)*8)



We calculate terms in parentheses: -((-9x)*8), so:

(-9x)*8

We multiply parentheses

-72x

Back to the equation:

-(-72x)



We get rid of parentheses

72x

Back to the equation:

+(72x)



We add all the numbers together, and all the variables

72x+(12x-20)/()=0

We multiply all the terms by the denominator


72x*()+(12x-20)=0

We get rid of parentheses

72x*()+12x-20=0

We add all the numbers together, and all the variables

12x+72x*()-20=0

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

12x+72x*()=20

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