7y+-1=2(1y+3)+-2

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Solution for 7y+-1=2(1y+3)+-2 equation: 



7y+-1=2(1y+3)+-2

We move all terms to the left:

7y+-1-(2(1y+3)+-2)=0

We add all the numbers together, and all the variables

7y-(2(y+3)+-2)-1+=0

We add all the numbers together, and all the variables

7y-(2(y+3)+-2)=0

We use the square of the difference formula

7y-(2(y+3)-2)=0



We calculate terms in parentheses: -(2(y+3)-2), so:

2(y+3)-2

We multiply parentheses

2y+6-2

We add all the numbers together, and all the variables

2y+4

Back to the equation:

-(2y+4)



We get rid of parentheses

7y-2y-4=0

We add all the numbers together, and all the variables

5y-4=0

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

5y=4

y=4/5

y=4/5

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