7y+3-3(3y-4)=50

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Solution for 7y+3-3(3y-4)=50 equation: 



7y+3-3(3y-4)=50

We move all terms to the left:

7y+3-3(3y-4)-(50)=0

We add all the numbers together, and all the variables

7y-3(3y-4)-47=0

We multiply parentheses

7y-9y+12-47=0

We add all the numbers together, and all the variables

-2y-35=0

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

-2y=35

y=35/-2

y=-17+1/2

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