<p>Can someone help solve this problem...</p>
<p>If 2|x+3|=4 and |y+1|/3=2, then |x+y| could equal each of the following EXCEPT
A) 0
B) 4
C) 8
D) 10
E) 12</p>
<p>the answer is D</p>
<p>2|x+3| = 4
|x+3| = 2
x+3 = 2
x+3 = -2
x = -1 or -5</p>
<p>|y+1|/3 = 2
|y+1| = 6
y + 1 = 6
y + 1 = -6
y = 5 or -7</p>
<p>|-1 + 5| = 4
|-1 + -7| = 8</p>
<p>|-5 + 5| = 0
|-5 + -7| = 12</p>
<p>The answer is D.</p>
<p>|x+3|=2 |y+1|=6
if x>-3 then x+3=2 x=-1
if x<-3 then x+3=-2 x=-5
if y>-1 then y+1=6 y=5
if y<-1 then y+1=-6 y=-7
you're now able to calculate all the possible |x+y|
and there's no way you can get a 10
perhaps there's a simpler way but i can't figure it out right now</p>
<p>thank you very much, it was much simpler than I thought. what about this one...</p>
<p>If x and y are positive numbers that satisfy the equation, what is the value of x/y. </p>
<p>square root of (x^2-t^2)=2t-x</p>
<p>it is a free-response question and the answer is 1.25</p>
<p>I did not feel it was necesary to create a new thread</p>
<p>where's y?</p>
<p>sorry, i meant x/t</p>
<p>hehe I'm not sure I have understood your problem
"square root of (x^2-t^2)=2t-x"
does it mean this
x^2-t^2=(2t-x)^2?</p>
<p>ok I got it x^2-t^2=4t^2-4tx+x^2
4tx=5t^2
4x=5t
x/t=1.25</p>