1/6(x+2)-1/9(2x-3)-(1/2(1-3x))=0

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Solution for 1/6(x+2)-1/9(2x-3)-(1/2(1-3x))=0 equation: 



1/6(x+2)-1/9(2x-3)-(1/2(1-3x))=0



Domain of the equation: 6(x+2)!=0

x∈R





Domain of the equation: 9(2x-3)!=0

x∈R





Domain of the equation: 2(1-3x))!=0

x∈R



We add all the numbers together, and all the variables

1/6(x+2)-1/9(2x-3)-(1/2(-3x+1))=0

We calculate fractions


(-12x^2x/(6(x+2)*9(2x-3)*2(-3x+1)))+(-54x^2x/(6(x+2)*9(2x-3)*2(-3x+1)))+(18x^22/(6(x+2)*9(2x-3)*2(-3x+1)))=0



We calculate terms in parentheses: +(-12x^2x/(6(x+2)*9(2x-3)*2(-3x+1))), so:

-12x^2x/(6(x+2)*9(2x-3)*2(-3x+1))

We multiply all the terms by the denominator


-12x^2x

Back to the equation:

+(-12x^2x)





We calculate terms in parentheses: +(-54x^2x/(6(x+2)*9(2x-3)*2(-3x+1))), so:

-54x^2x/(6(x+2)*9(2x-3)*2(-3x+1))

We multiply all the terms by the denominator


-54x^2x

Back to the equation:

+(-54x^2x)





We calculate terms in parentheses: +(18x^22/(6(x+2)*9(2x-3)*2(-3x+1))), so:

18x^22/(6(x+2)*9(2x-3)*2(-3x+1))

We do not support expression: x^22

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