3(b+-2)+5=7+3(2b+-5)+-38

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Solution for 3(b+-2)+5=7+3(2b+-5)+-38 equation: 



3(b+-2)+5=7+3(2b+-5)+-38

We move all terms to the left:

3(b+-2)+5-(7+3(2b+-5)+-38)=0

We add all the numbers together, and all the variables

3(b-2)-(7+3(2b-5)+-38)+5=0

We use the square of the difference formula

3(b-2)-(7+3(2b-5)-38)+5=0

We multiply parentheses

3b-(7+3(2b-5)-38)-6+5=0



We calculate terms in parentheses: -(7+3(2b-5)-38), so:

7+3(2b-5)-38

determiningTheFunctionDomain
3(2b-5)+7-38

We add all the numbers together, and all the variables

3(2b-5)-31

We multiply parentheses

6b-15-31

We add all the numbers together, and all the variables

6b-46

Back to the equation:

-(6b-46)



We add all the numbers together, and all the variables

3b-(6b-46)-1=0

We get rid of parentheses

3b-6b+46-1=0

We add all the numbers together, and all the variables

-3b+45=0

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

-3b=-45

b=-45/-3

b=+15

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