1/3(9c-18-3c)=1/4(12c-28)

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Solution for 1/3(9c-18-3c)=1/4(12c-28) equation: 



1/3(9c-18-3c)=1/4(12c-28)

We move all terms to the left:

1/3(9c-18-3c)-(1/4(12c-28))=0



Domain of the equation: 3(9c-18-3c)!=0

c∈R





Domain of the equation: 4(12c-28))!=0

c∈R



We add all the numbers together, and all the variables

1/3(6c-18)-(1/4(12c-28))=0

We calculate fractions


(4c1/(3(6c-18)*4(12c-28)))+(-3c6/(3(6c-18)*4(12c-28)))=0



We calculate terms in parentheses: +(4c1/(3(6c-18)*4(12c-28))), so:

4c1/(3(6c-18)*4(12c-28))

We multiply all the terms by the denominator


4c1

We add all the numbers together, and all the variables

4c

Back to the equation:

+(4c)





We calculate terms in parentheses: +(-3c6/(3(6c-18)*4(12c-28))), so:

-3c6/(3(6c-18)*4(12c-28))

We multiply all the terms by the denominator


-3c6

We add all the numbers together, and all the variables

-3c^6

We do not support ecpression: c^6

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