(8-4x)/3-2(5x+8)=(2(4x+6))/9+2(10x+1)

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Solution for (8-4x)/3-2(5x+8)=(2(4x+6))/9+2(10x+1) equation: 



(8-4x)/3-2(5x+8)=(2(4x+6))/9+2(10x+1)

We move all terms to the left:

(8-4x)/3-2(5x+8)-((2(4x+6))/9+2(10x+1))=0



Domain of the equation: 9+2(10x+1))!=0

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

2(10x+1))!=-9

x∈R



We add all the numbers together, and all the variables

(-4x+8)/3-2(5x+8)-((2(4x+6))/9+2(10x+1))=0

We multiply parentheses

(-4x+8)/3-10x-((2(4x+6))/9+2(10x+1))-16=0

We calculate fractions


-10x+(-16x)/(2(10x+1))-((2(4x+6))*3)/(2(10x+1))-16+3*9)+(+3*9)=0



We calculate terms in parentheses: +(-16x)/(2(10x+1)), so:

-16x)/(2(10x+1)

We multiply all the terms by the denominator


-16x)

Back to the equation:

+(-16x))





We calculate terms in parentheses: -((2(4x+6))*3)/(2(10x+1)), so:

(2(4x+6))*3)/(2(10x+1)

We multiply all the terms by the denominator


(2(4x+6))*3)



We calculate terms in parentheses: +(2(4x+6))*3), so:

2(4x+6))*3

We multiply parentheses

24x+

We add all the numbers together, and all the variables

24x

Back to the equation:

+(24x)



Back to the equation:

-(24x)



We add all the numbers together, and all the variables

-10x+(-16x))-24x-16+3*9)+27=0

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