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!!

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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!!

Mathematics
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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.
So \[-(3y+4)\]means\[-1 \times (3y+4)=-1 \times 3y +-1 \times 4=-3y-4\]
The second part is\[-(-2y-5)=-1 \times -2y+-1 \times -5=2y+5\]

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Other answers:

In the first expansion, we had (negative) x (positive) = (negative), and in the second, (negative) x (negative) = (positive).
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,\[y=1\]
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