(1/3)(12x-15)=(1/7)(35x-28)

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Solution for (1/3)(12x-15)=(1/7)(35x-28) equation: 



(1/3)(12x-15)=(1/7)(35x-28)

We move all terms to the left:

(1/3)(12x-15)-((1/7)(35x-28))=0



Domain of the equation: 3)(12x-15)!=0

x∈R





Domain of the equation: 7)(35x-28))!=0

x∈R



We add all the numbers together, and all the variables

(+1/3)(12x-15)-((+1/7)(35x-28))=0

We multiply parentheses ..

(+12x^2+1/3*-15)-((+1/7)(35x-28))=0

We calculate fractions


((+12x^2+1*7)(35x-28)))/(-15)*7)(35x-28))+3*)+()/(-15)*7)(35x-28))+3*)=0

We calculate fractions


(((+12x^2+1*7)(35x-28)))*(-15)*7)(35x-28))+3*))/((*(-15)*7)35x)+(-15)*7)(35x-28))+3*)+()*()/((*(-15)*7)35x)=0

We can not solve this equation

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