3(m-2)+m=4(m+3)-18

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Solution for 3(m-2)+m=4(m+3)-18 equation: 



3(m-2)+m=4(m+3)-18

We move all terms to the left:

3(m-2)+m-(4(m+3)-18)=0

We add all the numbers together, and all the variables

m+3(m-2)-(4(m+3)-18)=0

We multiply parentheses

m+3m-(4(m+3)-18)-6=0



We calculate terms in parentheses: -(4(m+3)-18), so:

4(m+3)-18

We multiply parentheses

4m+12-18

We add all the numbers together, and all the variables

4m-6

Back to the equation:

-(4m-6)



We add all the numbers together, and all the variables

4m-(4m-6)-6=0

We get rid of parentheses

4m-4m+6-6=0

We add all the numbers together, and all the variables

=0

m=0/1

m=0

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