12(x-2)+3x=(1/3)(x+12)+10x

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Solution for 12(x-2)+3x=(1/3)(x+12)+10x equation: 



12(x-2)+3x=(1/3)(x+12)+10x

We move all terms to the left:

12(x-2)+3x-((1/3)(x+12)+10x)=0



Domain of the equation: 3)(x+12)+10x)!=0

x∈R



We add all the numbers together, and all the variables

12(x-2)+3x-((+1/3)(x+12)+10x)=0

We add all the numbers together, and all the variables

3x+12(x-2)-((+1/3)(x+12)+10x)=0

We multiply parentheses

3x+12x-((+1/3)(x+12)+10x)-24=0

We multiply parentheses ..

-((+x^2+1/3*12)+10x)+3x+12x-24=0

We multiply all the terms by the denominator


-((+x^2+1+3x*3*12)+10x)+12x*3*12)+10x)-24*3*12)+10x)=0



We calculate terms in parentheses: -((+x^2+1+3x*3*12)+10x), so:

(+x^2+1+3x*3*12)+10x

We get rid of parentheses

x^2+3x*3*12+10x+1

We add all the numbers together, and all the variables

x^2+10x+3x*3*12+1

Wy multiply elements

x^2+10x+108x*1+1

Wy multiply elements

x^2+10x+108x+1

We add all the numbers together, and all the variables

x^2+118x+1

Back to the equation:

-(x^2+118x+1)



Wy multiply elements

15552x^3*1-(x^2+118x+1)=0

We get rid of parentheses

15552x^3*1-x^2-118x-1=0

We do not support expression: x^3

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