(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 multiply parentheses

8x+

We add all the numbers together, and all the variables

8x

Back to the equation:

-(8x)



We add all the numbers together, and all the variables

-10x+(-16x))-8x-16+3*9)+27=0

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