(3h+1)+(h-3)+1=7+(2h+4)

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Solution for (3h+1)+(h-3)+1=7+(2h+4) equation: 



(3h+1)+(h-3)+1=7+(2h+4)

We move all terms to the left:

(3h+1)+(h-3)+1-(7+(2h+4))=0

We get rid of parentheses

3h+h-(7+(2h+4))+1-3+1=0



We calculate terms in parentheses: -(7+(2h+4)), so:

7+(2h+4)

determiningTheFunctionDomain
(2h+4)+7

We get rid of parentheses

2h+4+7

We add all the numbers together, and all the variables

2h+11

Back to the equation:

-(2h+11)



We add all the numbers together, and all the variables

4h-(2h+11)-1=0

We get rid of parentheses

4h-2h-11-1=0

We add all the numbers together, and all the variables

2h-12=0

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

2h=12

h=12/2

h=6

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