3x-4+2x-1/2(2x-4)+2x-3=1/3(3x-27)+x-4+7x

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Solution for 3x-4+2x-1/2(2x-4)+2x-3=1/3(3x-27)+x-4+7x equation: 



3x-4+2x-1/2(2x-4)+2x-3=1/3(3x-27)+x-4+7x

We move all terms to the left:

3x-4+2x-1/2(2x-4)+2x-3-(1/3(3x-27)+x-4+7x)=0



Domain of the equation: 2(2x-4)!=0

x∈R





Domain of the equation: 3(3x-27)+x-4+7x)!=0

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

3(3x-27)+x+7x)!=4

x∈R



We add all the numbers together, and all the variables

7x-1/2(2x-4)-(1/3(3x-27)+x-4+7x)-7=0

We calculate fractions


(-21x^2-8x-4)/(x+2(2x-4)*3(3x-27)+7x)+7x-2x2/(x+2(2x-4)*3(3x-27)+7x)-7-4)+(-4)=0



We calculate terms in parentheses: +(-21x^2-8x-4)/(x+2(2x-4)*3(3x-27)+7x), so:

-21x^2-8x-4)/(x+2(2x-4)*3(3x-27)+7x

We add all the numbers together, and all the variables

-21x^2-1x-4)/(x+2(2x-4)*3(3x-27)

We multiply all the terms by the denominator


-21x^2*(x-1x*(x-4)+(2(2x-4)*3(3x-27))*(x

Back to the equation:

+(-21x^2*(x-1x*(x-4)+(2(2x-4)*3(3x-27))*(x)



We multiply all the terms by the denominator


(-21x^2*(x-1x*(x-4)+(2(2x-4)*3(3x-27))*x+7x-2x2=0

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