3y-7+4(y-6)+y-12+3(y+5)=2(y+9)+3y-10-y+55

Solution for 3y-7+4(y-6)+y-12+3(y+5)=2(y+9)+3y-10-y+55 equation:

3y-7+4(y-6)+y-12+3(y+5)=2(y+9)+3y-10-y+55

We move all terms to the left:

3y-7+4(y-6)+y-12+3(y+5)-(2(y+9)+3y-10-y+55)=0

We add all the numbers together, and all the variables

4y+4(y-6)+3(y+5)-(2(y+9)+3y-10-y+55)-19=0

We multiply parentheses

4y+4y+3y-(2(y+9)+3y-10-y+55)-24+15-19=0


We calculate terms in parentheses: -(2(y+9)+3y-10-y+55), so:

2(y+9)+3y-10-y+55

determiningTheFunctionDomain
2(y+9)+3y-y-10+55

We add all the numbers together, and all the variables

2y+2(y+9)+45

We multiply parentheses

2y+2y+18+45

We add all the numbers together, and all the variables

4y+63

Back to the equation:

-(4y+63)


We add all the numbers together, and all the variables

11y-(4y+63)-28=0

We get rid of parentheses

11y-4y-63-28=0

We add all the numbers together, and all the variables

7y-91=0

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

7y=91

y=91/7

y=13