11(y+3)=4(y-12)+25

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Solution for 11(y+3)=4(y-12)+25 equation: 



11(y+3)=4(y-12)+25

We move all terms to the left:

11(y+3)-(4(y-12)+25)=0

We multiply parentheses

11y-(4(y-12)+25)+33=0



We calculate terms in parentheses: -(4(y-12)+25), so:

4(y-12)+25

We multiply parentheses

4y-48+25

We add all the numbers together, and all the variables

4y-23

Back to the equation:

-(4y-23)



We get rid of parentheses

11y-4y+23+33=0

We add all the numbers together, and all the variables

7y+56=0

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

7y=-56

y=-56/7

y=-8

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