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

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



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

We move all terms to the left:

-(y+3)+(y-4)-((3+y)-(4-y))=0

We add all the numbers together, and all the variables

-(y+3)+(y-4)-((y+3)-(-1y+4))=0

We get rid of parentheses

-y+y-((y+3)-(-1y+4))-3-4=0



We calculate terms in parentheses: -((y+3)-(-1y+4)), so:

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

We get rid of parentheses

y+1y+3-4

We add all the numbers together, and all the variables

2y-1

Back to the equation:

-(2y-1)



We add all the numbers together, and all the variables

-(2y-1)-7=0

We get rid of parentheses

-2y+1-7=0

We add all the numbers together, and all the variables

-2y-6=0

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

-2y=6

y=6/-2

y=-3

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