(1)/(3)(6x+24)=-(1)/(4)(12x-72)+20

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Solution for (1)/(3)(6x+24)=-(1)/(4)(12x-72)+20 equation: 



(1)/(3)(6x+24)=-(1)/(4)(12x-72)+20

We move all terms to the left:

(1)/(3)(6x+24)-(-(1)/(4)(12x-72)+20)=0



Domain of the equation: 3(6x+24)!=0

x∈R





Domain of the equation: 4(12x-72)+20)!=0

x∈R



We calculate fractions


(4x1/(3(6x+24)*4(12x-72))+(-(-3x6)/(3(6x+24)*4(12x-72))=0



We calculate terms in parentheses: +(4x1/(3(6x+24)*4(12x-72))+(-(-3x6)/(3(6x+24)*4(12x-72)), so:

4x1/(3(6x+24)*4(12x-72))+(-(-3x6)/(3(6x+24)*4(12x-72)

We add all the numbers together, and all the variables

(-(-3x^6)/(3(6x+24)*4(12x-72)+4x1/(3(6x+24)*4(12x-72))

We can not solve this equation

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