x+(2/3x+300)+(7/6x-10)=1990

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Solution for x+(2/3x+300)+(7/6x-10)=1990 equation: 



x+(2/3x+300)+(7/6x-10)=1990

We move all terms to the left:

x+(2/3x+300)+(7/6x-10)-(1990)=0



Domain of the equation: 3x+300)!=0

x∈R





Domain of the equation: 6x-10)!=0

x∈R



We get rid of parentheses

x+2/3x+7/6x+300-10-1990=0

We calculate fractions


x+12x/18x^2+21x/18x^2+300-10-1990=0

We add all the numbers together, and all the variables

x+12x/18x^2+21x/18x^2-1700=0

We multiply all the terms by the denominator


x*18x^2+12x+21x-1700*18x^2=0

We add all the numbers together, and all the variables

33x+x*18x^2-1700*18x^2=0

Wy multiply elements

18x^3-30600x^2+33x=0

We do not support expression: x^3

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