9x+(1/2)=(2/3)x-8

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Solution for 9x+(1/2)=(2/3)x-8 equation:



9x+(1/2)=(2/3)x-8
We move all terms to the left:
9x+(1/2)-((2/3)x-8)=0
Domain of the equation: 3)x-8)!=0
x!=0/1
x!=0
x∈R
We add all the numbers together, and all the variables
9x-((+2/3)x-8)+(+1/2)=0
We get rid of parentheses
9x-((+2/3)x-8)+1/2=0
We calculate fractions
9x+()/3x-8)*2)+3x/3x-8)*2)=0
We calculate fractions
9x+(()*3x)/9x^2+(-8)*2)+3x*3x)/9x^2=0
We multiply all the terms by the denominator
9x*9x^2+(()*3x)+(-8)*2)+3x*3x)=0
We calculate terms in parentheses: +(()*3x), so:
()*3x
Wy multiply elements
81x^3+(()*3x)+(-8)*2)+3x*3x)=0
We get rid of parentheses
81x^3+(()*3x)-8)*2)+3x*3x=0
We calculate terms in parentheses: +(()*3x), so:
()*3x
We do not support expression: x^3

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