48-(3+10y-4(y+2))=-3(5y-4)-(9(y-1)-14y+4)

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Solution for 48-(3+10y-4(y+2))=-3(5y-4)-(9(y-1)-14y+4) equation: 



48-3+10y-4(y+2)=-3(5y-4)-9(y-1)-14y+4

We move all terms to the left:

48-3+10y-4(y+2)-(-3(5y-4)-9(y-1)-14y+4)=0

We add all the numbers together, and all the variables

10y-4(y+2)-(-3(5y-4)-9(y-1)-14y+4)+45=0

We multiply parentheses

10y-4y-(-3(5y-4)-9(y-1)-14y+4)-8+45=0



We calculate terms in parentheses: -(-3(5y-4)-9(y-1)-14y+4), so:

-3(5y-4)-9(y-1)-14y+4

We add all the numbers together, and all the variables

-14y-3(5y-4)-9(y-1)+4

We multiply parentheses

-14y-15y-9y+12+9+4

We add all the numbers together, and all the variables

-38y+25

Back to the equation:

-(-38y+25)



We add all the numbers together, and all the variables

6y-(-38y+25)+37=0

We get rid of parentheses

6y+38y-25+37=0

We add all the numbers together, and all the variables

44y+12=0

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

44y=-12

y=-12/44

y=-3/11

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