## anonymous 5 years ago I'm having trouble with these types of problems: -(3y+4)-(-2y-5)=0. I don't know when I'm supposed to change the negative to positive or vice versa. Someone please me!!

1. anonymous

The rules for multiplying positive and negative numbers are the following:$2 \times 3 = 6$$2 \times -3 = -6$$-2 \times 3 = -6$$-2 \times - 3 = 6$ So you should see, (positive) x (positive) = (positive) (positive) x (negative) = (negative) (negative) x (positive) = (negative) (negative) x (negative) = (positive) If you have two same signs when multiplying, the number is positive. If you have two different signs when multiplying, the number is negative.

2. anonymous

So $-(3y+4)$means$-1 \times (3y+4)=-1 \times 3y +-1 \times 4=-3y-4$

3. anonymous

The second part is$-(-2y-5)=-1 \times -2y+-1 \times -5=2y+5$

4. anonymous

In the first expansion, we had (negative) x (positive) = (negative), and in the second, (negative) x (negative) = (positive).

5. anonymous

Combine them to give,$-3y-4+2y+5=0$which looks like$-y+1=0$after collecting like-terms. Adding y to both sides gives you,$y=1$

6. anonymous

Does this help?