1/3(2x-4)+5=2/3(x+1)=

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



1/3(2x-4)+5=2/3(x+1)=

We move all terms to the left:

1/3(2x-4)+5-(2/3(x+1))=0



Domain of the equation: 3(2x-4)!=0

x∈R





Domain of the equation: 3(x+1))!=0

x∈R



We calculate fractions


(3xx/(3(2x-4)*3(x+1)))+(-6x2/(3(2x-4)*3(x+1)))+5=0



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

3xx/(3(2x-4)*3(x+1))

We multiply all the terms by the denominator


3xx

Back to the equation:

+(3xx)





We calculate terms in parentheses: +(-6x2/(3(2x-4)*3(x+1))), so:

-6x2/(3(2x-4)*3(x+1))

We multiply all the terms by the denominator


-6x2

We add all the numbers together, and all the variables

-6x^2

Back to the equation:

+(-6x^2)



We get rid of parentheses

-6x^2+3xx+5=0

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

-6x^2+3xx=-5

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