1/5(3t-4)=1/3(t+12)

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Solution for 1/5(3t-4)=1/3(t+12) equation: 



1/5(3t-4)=1/3(t+12)

We move all terms to the left:

1/5(3t-4)-(1/3(t+12))=0



Domain of the equation: 5(3t-4)!=0

t∈R





Domain of the equation: 3(t+12))!=0

t∈R



We calculate fractions


(3tt/(5(3t-4)*3(t+12)))+(-5t3/(5(3t-4)*3(t+12)))=0



We calculate terms in parentheses: +(3tt/(5(3t-4)*3(t+12))), so:

3tt/(5(3t-4)*3(t+12))

We multiply all the terms by the denominator


3tt

Back to the equation:

+(3tt)





We calculate terms in parentheses: +(-5t3/(5(3t-4)*3(t+12))), so:

-5t3/(5(3t-4)*3(t+12))

We multiply all the terms by the denominator


-5t3

We add all the numbers together, and all the variables

-5t^3

We do not support etpression: t^3

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