anonymous
  • anonymous
if y<-2, then |y+2|/(y+2)=? i do not get this at all can someone please explain this to me?? what am I suppose to do and why?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
\[|y+2|=y+2 \text{ if } y+2>0 => y>-2\] \[|y+2|=-(y+2) \text{ if } y+2<0 => y<-2\]
myininaya
  • myininaya
\[\frac{-(y+2)}{y+2} =\frac{-1}{1}=-1\]
amistre64
  • amistre64
since its an inequality your output is a range of values

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amistre64
  • amistre64
top is always + and bottom always - so the results are going to be - as a result
anonymous
  • anonymous
okay so what would you do if 14
amistre64
  • amistre64
x is always grater than 14
amistre64
  • amistre64
14-14 = 0 and everything gerter than that is just +
amistre64
  • amistre64
greater than that
amistre64
  • amistre64
...negative, but still the abs makes it positives
anonymous
  • anonymous
I know but the solution is -14+x why did they choose to do -(14-x) instead of 14-x
anonymous
  • anonymous
I must be missing something....
amistre64
  • amistre64
say x is greater than 14, x=20 for example: 14-20=-6 but we cant HAVE a -6 equal to an | |
myininaya
  • myininaya
\[|14-x| =14-x \text{ if } 14-x>0\] \[|14-x|=-(14-x) \text{ if } 14-x<0\]
amistre64
  • amistre64
so they chose the one that will match it
myininaya
  • myininaya
we want the output to be postive
myininaya
  • myininaya
if 14-x<0, then -(14-x)>-0
myininaya
  • myininaya
that is 0 not -0
anonymous
  • anonymous
ahhh so y<-2 we want the output to be negative?
amistre64
  • amistre64
yes
myininaya
  • myininaya
\[y<-2 => y+2<0 => |y+2|=-(y+2) \text{ since y+2 is negative }\]
anonymous
  • anonymous
I think I get it now that you guys so much!!! you just kind of have to sample the equation
amistre64
  • amistre64
yes, sample it to see what will make sense overall :)
myininaya
  • myininaya
i like my way if x>0 then |x|=x if x<0 then |x|=-x (-x gives us postive output because x is negative)
myininaya
  • myininaya
x>0 means x is positive while x<0 means x is negative

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