16(y+2)=3(y-10)+49

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Solution for 16(y+2)=3(y-10)+49 equation: 



16(y+2)=3(y-10)+49

We move all terms to the left:

16(y+2)-(3(y-10)+49)=0

We multiply parentheses

16y-(3(y-10)+49)+32=0



We calculate terms in parentheses: -(3(y-10)+49), so:

3(y-10)+49

We multiply parentheses

3y-30+49

We add all the numbers together, and all the variables

3y+19

Back to the equation:

-(3y+19)



We get rid of parentheses

16y-3y-19+32=0

We add all the numbers together, and all the variables

13y+13=0

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

13y=-13

y=-13/13

y=-1

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