7(y+3)=-3(4y-2)+y

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Solution for 7(y+3)=-3(4y-2)+y equation:



7(y+3)=-3(4y-2)+y
We move all terms to the left:
7(y+3)-(-3(4y-2)+y)=0
We multiply parentheses
7y-(-3(4y-2)+y)+21=0
We calculate terms in parentheses: -(-3(4y-2)+y), so:
-3(4y-2)+y
We add all the numbers together, and all the variables
y-3(4y-2)
We multiply parentheses
y-12y+6
We add all the numbers together, and all the variables
-11y+6
Back to the equation:
-(-11y+6)
We get rid of parentheses
7y+11y-6+21=0
We add all the numbers together, and all the variables
18y+15=0
We move all terms containing y to the left, all other terms to the right
18y=-15
y=-15/18
y=-5/6

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