14-1/3(f-10)=2/3(25+f)

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Solution for 14-1/3(f-10)=2/3(25+f) equation: 



14-1/3(f-10)=2/3(25+f)

We move all terms to the left:

14-1/3(f-10)-(2/3(25+f))=0



Domain of the equation: 3(f-10)!=0

f∈R





Domain of the equation: 3(25+f))!=0

f∈R



We add all the numbers together, and all the variables

-1/3(f-10)-(2/3(f+25))+14=0

We calculate fractions


(-3ff/(3(f-10)*3(f+25)))+(-6ff/(3(f-10)*3(f+25)))+14=0



We calculate terms in parentheses: +(-3ff/(3(f-10)*3(f+25))), so:

-3ff/(3(f-10)*3(f+25))

We multiply all the terms by the denominator


-3ff

Back to the equation:

+(-3ff)





We calculate terms in parentheses: +(-6ff/(3(f-10)*3(f+25))), so:

-6ff/(3(f-10)*3(f+25))

We multiply all the terms by the denominator


-6ff

Back to the equation:

+(-6ff)



We get rid of parentheses

-3ff-6ff+14=0

We move all terms containing f to the left, all other terms to the right

-3ff-6ff=-14

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