(4y)+(y+1)+(3y+1)=68

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Solution for (4y)+(y+1)+(3y+1)=68 equation: 



(4y)+(y+1)+(3y+1)=68

We move all terms to the left:

(4y)+(y+1)+(3y+1)-(68)=0

We get rid of parentheses

4y+y+3y+1+1-68=0

We add all the numbers together, and all the variables

8y-66=0

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

8y=66

y=66/8

y=8+1/4

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