5=(2/3)(2)+b

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Solution for 5=(2/3)(2)+b equation: 



5=(2/3)(2)+b

We move all terms to the left:

5-((2/3)(2)+b)=0



Domain of the equation: 3)2+b)!=0

b!=0/1

b!=0

b∈R



We add all the numbers together, and all the variables

-((+2/3)2+b)+5=0

We multiply all the terms by the denominator


-((+2+5*3)2+b)=0



We calculate terms in parentheses: -((+2+5*3)2+b), so:

(+2+5*3)2+b

determiningTheFunctionDomain
b+(+2+5*3)2

We add all the numbers together, and all the variables

b+172

Back to the equation:

-(b+172)



We get rid of parentheses

-b-172=0

We add all the numbers together, and all the variables

-1b-172=0

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

-b=172

b=172/-1

b=-172

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