2/3(x+2)-1/2(x-4)=1/4(x+5)

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Solution for 2/3(x+2)-1/2(x-4)=1/4(x+5) equation: 



2/3(x+2)-1/2(x-4)=1/4(x+5)

We move all terms to the left:

2/3(x+2)-1/2(x-4)-(1/4(x+5))=0



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

x∈R





Domain of the equation: 2(x-4)!=0

x∈R





Domain of the equation: 4(x+5))!=0

x∈R



We calculate fractions


(-12x^2x/(3(x+2)*2(x-4)*4(x+5)))+(-6x^2x/(3(x+2)*2(x-4)*4(x+5)))+(16x^2x/(3(x+2)*2(x-4)*4(x+5)))=0



We calculate terms in parentheses: +(-12x^2x/(3(x+2)*2(x-4)*4(x+5))), so:

-12x^2x/(3(x+2)*2(x-4)*4(x+5))

We multiply all the terms by the denominator


-12x^2x

Back to the equation:

+(-12x^2x)





We calculate terms in parentheses: +(-6x^2x/(3(x+2)*2(x-4)*4(x+5))), so:

-6x^2x/(3(x+2)*2(x-4)*4(x+5))

We multiply all the terms by the denominator


-6x^2x

Back to the equation:

+(-6x^2x)





We calculate terms in parentheses: +(16x^2x/(3(x+2)*2(x-4)*4(x+5))), so:

16x^2x/(3(x+2)*2(x-4)*4(x+5))

We multiply all the terms by the denominator


16x^2x

Back to the equation:

+(16x^2x)



We get rid of parentheses

-12x^2x-6x^2x+16x^2x=0

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