<p>i got it right, but it was from guessing..can u tell me how to really solve it?</p>
<p>If 2|x+3|=4 and |y+1|/3=2, then |x+y| could equal each of the following EXCEPT</p>
<p>(A) 0
(B) 4
(C) 8
(D) 10
(E) 12</p>
<p>thx</p>
<p>answer is 8 isnt it??</p>
<p>^no, it's 10</p>
<p>You have to look at each seperately. Take 2|x+3|=4 first</p>
<p>2|x+3|=4 </p>
<p>(divide both sides by two)</p>
<p>|x+3|=2</p>
<p>(now you know that there are two cases: one where X+3 is pos and one where X+3 is neg; therefore you write two out)</p>
<p>
[quote]
CASE 1
x+3=2
x=-1
[/quote]
</p>
<p>
[quote]
CASE 2
-|x+3|=2
-x-3=2
-x=5
x=-5
[/quote]
</p>
<p>Now you have two values of X. Do the same thing for y, and you get two values (y=-7, 5)</p>
<p>Then you look at the combination that |x+y| can give you; remember the two values of each (x=-5,-1; y=-7,5). The only value you can't get is 10</p>
<p>I second that. The answer is D.
Out of those absolute values you can get 4 values of x and y (2 per each). And 4 possible values for x+y. 10 is not any of these values.</p>
<p>k i c now...i narrowed it down to a and d (0 and 10) and just guessed 10 cuz it looked better and it was the only answer that wasnt like a factor of 4 (well 0, then +4=4, then +4=8, then +4=12) so 10 looked like a good answer..lol</p>
<p>alright i c now i forgot to take the negative of the absolute value...thanks for the help!!</p>
<p>y= -7 or 5
x= -1 or -5
add them up and the only one not there is 10 (D)</p>