0=(1r+2)-(4+2r)

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Solution for 0=(1r+2)-(4+2r) equation: 



0=(1r+2)-(4+2r)

We move all terms to the left:

0-((1r+2)-(4+2r))=0

We add all the numbers together, and all the variables

-((r+2)-(2r+4))+0=0

We add all the numbers together, and all the variables

-((r+2)-(2r+4))=0



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

(r+2)-(2r+4)

We get rid of parentheses

r-2r+2-4

We add all the numbers together, and all the variables

-1r-2

Back to the equation:

-(-1r-2)



We get rid of parentheses

1r+2=0

We add all the numbers together, and all the variables

r+2=0

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

r=-2

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