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Ds qusetion

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Ds qusetion

by jamesk486 » Wed May 23, 2007 11:38 am
If y>0, what is the value of x?
1. |x-3|>y
2. |x-3|<-y

i'm a bit confused about condition 2 bc even if |x-3| must be positive, how can it be less than -y which is negative since y>0 ?
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Re: Ds qusetion

by gabriel » Thu May 24, 2007 9:44 am
jamesk486 wrote:If y>0, what is the value of x?
1. |x-3|>y
2. |x-3|<-y

i'm a bit confused about condition 2 bc even if |x-3| must be positive, how can it be less than -y which is negative since y>0 ?

chk the question again ... i bet it says y is > or equal to 0 ..
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by Cybermusings » Fri May 25, 2007 9:37 am
This question is not posted correctly

We already know that y>0....hence -y would be negative....now Ix-3I will always be positive....So how can Ix-3I be ever less than -y???
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by f2001290 » Fri May 25, 2007 10:56 am
Second statement is <=
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