40=-2/3(-12-12r)

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Solution for 40=-2/3(-12-12r) equation: 



40=-2/3(-12-12r)

We move all terms to the left:

40-(-2/3(-12-12r))=0



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

r∈R



We add all the numbers together, and all the variables

-(-2/3(-12r-12))+40=0

We multiply all the terms by the denominator


-(-2+40*3(-12r-12))=0



We calculate terms in parentheses: -(-2+40*3(-12r-12)), so:

-2+40*3(-12r-12)

determiningTheFunctionDomain
40*3(-12r-12)-2

Wy multiply elements

120r(--2

We use the square of the difference formula

120r(+2

Back to the equation:

-(120r(+2)



We add all the numbers together, and all the variables

-(120r2=0

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