2(m-3)+(m+1)=10-(m-1)

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Solution for 2(m-3)+(m+1)=10-(m-1) equation: 



2(m-3)+(m+1)=10-(m-1)

We move all terms to the left:

2(m-3)+(m+1)-(10-(m-1))=0

We multiply parentheses

2m+(m+1)-(10-(m-1))-6=0

We get rid of parentheses

2m+m-(10-(m-1))+1-6=0



We calculate terms in parentheses: -(10-(m-1)), so:

10-(m-1)

determiningTheFunctionDomain
-(m-1)+10

We get rid of parentheses

-m+1+10

We add all the numbers together, and all the variables

-1m+11

Back to the equation:

-(-1m+11)



We add all the numbers together, and all the variables

3m-(-1m+11)-5=0

We get rid of parentheses

3m+1m-11-5=0

We add all the numbers together, and all the variables

4m-16=0

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

4m=16

m=16/4

m=4

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