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!!

Question

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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,$$y=1$$

anonymous · Answer

Does this help?