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anonymous
 5 years ago
I'm having trouble with these types of problems: (3y+4)(2y5)=0. I don't know when I'm supposed to change the negative to positive or vice versa. Someone please me!!
anonymous
 5 years ago
I'm having trouble with these types of problems: (3y+4)(2y5)=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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0So \[(3y+4)\]means\[1 \times (3y+4)=1 \times 3y +1 \times 4=3y4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The second part is\[(2y5)=1 \times 2y+1 \times 5=2y+5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the first expansion, we had (negative) x (positive) = (negative), and in the second, (negative) x (negative) = (positive).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Combine them to give,\[3y4+2y+5=0\]which looks like\[y+1=0\]after collecting liketerms. Adding y to both sides gives you,\[y=1\]
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