3-(y-2)=2(y+1)-3(y-1)

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



3-(y-2)=2(y+1)-3(y-1)
We move all terms to the left:
3-(y-2)-(2(y+1)-3(y-1))=0
We get rid of parentheses
-y-(2(y+1)-3(y-1))+2+3=0
We calculate terms in parentheses: -(2(y+1)-3(y-1)), so:
2(y+1)-3(y-1)
We multiply parentheses
2y-3y+2+3
We add all the numbers together, and all the variables
-1y+5
Back to the equation:
-(-1y+5)
We add all the numbers together, and all the variables
-1y-(-1y+5)+5=0
We get rid of parentheses
-1y+1y-5+5=0
We add all the numbers together, and all the variables
=0
y=0/1
y=0

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