(4y+3)+(3y+7)=(9y-2)

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Solution for (4y+3)+(3y+7)=(9y-2) equation: 



(4y+3)+(3y+7)=(9y-2)

We move all terms to the left:

(4y+3)+(3y+7)-((9y-2))=0

We get rid of parentheses

4y+3y-((9y-2))+3+7=0



We calculate terms in parentheses: -((9y-2)), so:

(9y-2)

We get rid of parentheses

9y-2

Back to the equation:

-(9y-2)



We add all the numbers together, and all the variables

7y-(9y-2)+10=0

We get rid of parentheses

7y-9y+2+10=0

We add all the numbers together, and all the variables

-2y+12=0

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

-2y=-12

y=-12/-2

y=+6

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