4(3y+1)=6(2y-1)+10

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Solution for 4(3y+1)=6(2y-1)+10 equation: 



4(3y+1)=6(2y-1)+10

We move all terms to the left:

4(3y+1)-(6(2y-1)+10)=0

We multiply parentheses

12y-(6(2y-1)+10)+4=0



We calculate terms in parentheses: -(6(2y-1)+10), so:

6(2y-1)+10

We multiply parentheses

12y-6+10

We add all the numbers together, and all the variables

12y+4

Back to the equation:

-(12y+4)



We get rid of parentheses

12y-12y-4+4=0

We add all the numbers together, and all the variables

=0

y=0/1

y=0

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