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Re: If |p−6|+p=6, which of the following must be true? [#permalink]
30 Jun 2013, 17:08

If |p−6|+p=6, which of the following must be true?

There are two ways to solve, one of which is to just pick numbers.

The other way: |p-6|+p=6 |p-6| = 6-p |p-6| = -(p-6) |x| = -x (i.e. the value inside the absolute value bars is negative) So, (p-6) is negative when p≤6

Re: If |p−6|+p=6, which of the following must be true? [#permalink]
01 Jul 2013, 13:39

without plugging in numbers, how would you know that this |p-6| = -(p-6) would be p<=6 and not just p<6? the way i'm solving it, this |x| = -x condition would be met when (p-6)<0 and therefore -(p-6)<0 yields p<6. but why also p=6?

Re: If |p−6|+p=6, which of the following must be true? [#permalink]
01 Jul 2013, 13:42

Expert's post

nancerella wrote:

without plugging in numbers, how would you know that this |p-6| = -(p-6) would be p<=6 and not just p<6? the way i'm solving it, this |x| = -x condition would be met when (p-6)<0 and therefore -(p-6)<0 yields p<6. but why also p=6?

thanks!

Because when x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}.

We have |p-6|=-(p-6), thus p-6\leq{0} --> p\leq{6}.