SOLUTION: I am trying to help my 9th grader with algebra and I haven't a clue...She gets confused with the negative/positive part. When do you change or not: this is the problem: 3(5y+2)=4

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: I am trying to help my 9th grader with algebra and I haven't a clue...She gets confused with the negative/positive part. When do you change or not: this is the problem: 3(5y+2)=4      Log On


   

Question 20045: I am trying to help my 9th grader with algebra and I haven't a clue...She gets confused with the negative/positive part. When do you change or not: this is the problem: 3(5y+2)=4(3y+2)-1
PLEASE help. thanks dw
Answer by elima(1433) About Me  (Show Source):
You can put this solution on YOUR website!
3(5y+2)=4(3y+2)-1
First distribute the 3 and the 4;
15y+6=12y+8-1
15y+6=12y+7
Get the variable to one side; so you will subtract the 12y from both sides;
15y+6-12y=12y+7-12y
Now you have;
3y+6=7, now move the 6 to the other side;
3y+6-6=7-6
3y=1
Divide each side by 3;
3y%2F3=1%2F3
y=1%2F3
The rule for when to use negative or positive would;
You do the OPPOSITE OPERATION!
So if you have a problem such as;
7x+3=-4
you will subtract 3 from both sides because it is being added.
7x+3-3=-4-3
In the case of;
3x-5=16
You would add 5 to both sides because it is being subtracted.
3x-5+5=16+5
That is what I mean by the opposite operation. In the first case the operation was addition; the opposite would be subtraction, in the second case the operation was subtraction so the opposite would be addition!
Hope you understand!
If not let me know!
=)