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Mathematics 45 Online
OpenStudy (anonymous):

The number of solutions that are possible for the inequality \[\left[ x-5 \right] +\left| x -1\right|<2\] is

OpenStudy (experimentx):

[x-] ??

OpenStudy (anonymous):

sorry it is -5

OpenStudy (experimentx):

and what is [] <--- is it floor or ceil??

OpenStudy (experimentx):

or absolute value??

OpenStudy (anonymous):

absolute value

OpenStudy (unklerhaukus):

you ment this right/? \[|x−5|+|x−1|<2\]

OpenStudy (experimentx):

no solution http://www.wolframalpha.com/input/?i= |x-5|+%2B|x-1|++-+2

OpenStudy (anonymous):

yes u r correct 0 solution

OpenStudy (anonymous):

but how?

OpenStudy (experimentx):

is |x-5| + |x-1| ever less than zero??

OpenStudy (experimentx):

sorry .. ever less than 2?? or |x-5| + |x-1| - 2 less than zero ... check the graph

OpenStudy (anonymous):

so...... the sign should be > to have solution is it...

OpenStudy (experimentx):

No ... you just need to find a point for which the above in equality is true eg 3 < x < 4 3.5 is greater than 3 and less than 4 ... so this is true ... so this is solution

OpenStudy (anonymous):

can u explain it more..... plzzzz

OpenStudy (experimentx):

You just need to find the numbers that satisfy the given condition.

OpenStudy (experimentx):

i mean you have to find that numbers for which the given condition is true. for eg: 1 > x > 2 (find a number that is less than and greater than two ... find it's solution

OpenStudy (anonymous):

plzzz help !

OpenStudy (anonymous):

i did nt understand

OpenStudy (anonymous):

plzz help any one plz!!

OpenStudy (anonymous):

See the attached graph to see that |x-5| + | x-1| is always bigger than 2. There are no solutions. If this does not convince you try to consider three cases x <1 1< x < 5 x > 5 In each case try to solve it. I will do the first case for you if x< 1 the inequality becomes 5-x + 1 -x < 2 -2 x < -4 x > 2 contradiction with the face that x<1 Mimic for the remaining cases and conclude.

OpenStudy (anonymous):

@eliassaab the next i got x<4

OpenStudy (anonymous):

1 < x < 5 |x-5| + | x-1| < 2 5- x + x -1 < 2 4 < 2 contradiction. Do the last case.

OpenStudy (anonymous):

i f x > 5 |x-5| + | x-1| < 2 x-5 + x -1 < 2 2 x < 8 x < 4 but x > 5 contradiction.

OpenStudy (anonymous):

so......

OpenStudy (anonymous):

Did you understand it?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

then ................

OpenStudy (anonymous):

You are done. You showed that all cases lead to contradictions. So there are no solutions. Did you get it now?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

yw

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