16=1/24(3+b2)

Solution for 16=1/24(3+b2) equation:

16=1/24(3+b2)

We move all terms to the left:

16-(1/24(3+b2))=0


Domain of the equation: 24(3+b2))!=0

b∈R


We add all the numbers together, and all the variables

-(1/24(+b^2+3))+16=0

We multiply all the terms by the denominator


-(1+16*24(+b^2+3))=0


We calculate terms in parentheses: -(1+16*24(+b^2+3)), so:

1+16*24(+b^2+3)

determiningTheFunctionDomain
16*24(+b^2+3)+1

Wy multiply elements

384b(++1

We use the square of the difference formula

384b(+1

Back to the equation:

-(384b(+1)


We add all the numbers together, and all the variables

-(384b1=0