140=(3y+1)+(3y+1)+(4y-1)+(4y-1)

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Solution for 140=(3y+1)+(3y+1)+(4y-1)+(4y-1) equation:



140=(3y+1)+(3y+1)+(4y-1)+(4y-1)
We move all terms to the left:
140-((3y+1)+(3y+1)+(4y-1)+(4y-1))=0
We calculate terms in parentheses: -((3y+1)+(3y+1)+(4y-1)+(4y-1)), so:
(3y+1)+(3y+1)+(4y-1)+(4y-1)
We get rid of parentheses
3y+3y+4y+4y+1+1-1-1
We add all the numbers together, and all the variables
14y
Back to the equation:
-(14y)
We move all terms containing y to the left, all other terms to the right
-14y=-140
y=-140/-14
y=+10

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