(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 add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


-(-6x^2)

We get rid of parentheses

6x^2

Back to the equation:

+(6x^2)



determiningTheFunctionDomain
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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