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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