1/3(x-2)+1/4(x+3)=1/5(x+4)+15

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Solution for 1/3(x-2)+1/4(x+3)=1/5(x+4)+15 equation: 



1/3(x-2)+1/4(x+3)=1/5(x+4)+15

We move all terms to the left:

1/3(x-2)+1/4(x+3)-(1/5(x+4)+15)=0



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

x∈R





Domain of the equation: 4(x+3)!=0

x∈R





Domain of the equation: 5(x+4)+15)!=0

x∈R



We calculate fractions


(20x^2x/(3(x-2)*4(x+3)*5(x+4))+(15x^2x/(3(x-2)*4(x+3)*5(x+4))+(-12x^2x/(3(x-2)*4(x+3)*5(x+4))=0



We calculate terms in parentheses: +(20x^2x/(3(x-2)*4(x+3)*5(x+4))+(15x^2x/(3(x-2)*4(x+3)*5(x+4))+(-12x^2x/(3(x-2)*4(x+3)*5(x+4)), so:

20x^2x/(3(x-2)*4(x+3)*5(x+4))+(15x^2x/(3(x-2)*4(x+3)*5(x+4))+(-12x^2x/(3(x-2)*4(x+3)*5(x+4)

We can not solve this equation

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