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PAT inequality question

https://oxfordpat.wordpress.com/oxford-pat-2016-question-9/

on the answer, why is it 0 < x <2/3 rather than just x < 2/3
Well firstly x cannot be zero and if x was negative then the left hand side would also be bigger than right hand side.
Original post by alvan15
https://oxfordpat.wordpress.com/oxford-pat-2016-question-9/

on the answer, why is it 0 < x <2/3 rather than just x < 2/3


You can't allow x = 0, also looking at where the function is > 0 only applies to x > 0, once you look at x < 0 you have to switch to the 2nd inequality, which has no solutions. So the only good bits are x > 1 and 0 < x < 2/3.
(edited 6 years ago)
Reply 3
Original post by Physics Enemy
You can't allow x = 0, also looking at where the function is > 0 only applies to x > 0, once you look at x < 0 you have to switch to the 2nd inequality, which has no solutions. So the only good bits are x > 1 and that small region between 0 and 2/3.


by 2nd inequality do you mean when the inequality sign is flipped?
Reply 4
Original post by Physics Enemy
You can't allow x = 0, also looking at where the function is > 0 only applies to x > 0, once you look at x < 0 you have to switch to the 2nd inequality, which has no solutions. So the only good bits are x > 1 and 0 < x < 2/3.


ah i see, so i have to be careful when i turn the inequality equation in to a quadratic equation by times x
Original post by alvan15
by 2nd inequality do you mean when the inequality sign is flipped?


Yes, when x flips to < 0 you have to look at where the function is < 0, which it never is. You were still looking at the 1st inequality for x > 0 (and forgot x =/= 0).
Reply 6
Original post by Physics Enemy
Yes, when x flips to < 0 you have to look at where the function is < 0, which it never is. You were still looking at the 1st inequality for x > 0 (and forgot x =/= 0).


cheers , makes totally sense now
Original post by alvan15
ah i see, so i have to be careful when i turn the inequality equation in to a quadratic equation by times x


That care is needed to form 2 inequalities instead of 1 (for x > 0 and x < 0). The next bit of care is in using these together with the graph, to find your solutions.
Reply 8
Original post by Physics Enemy
That care is needed to form 2 inequalities instead of 1 (for x > 0 and x < 0). The next bit of care is in using these together with the graph, to find your solutions.


just want to say thank you again for helping :smile:
Original post by alvan15
just want to say thank you again for helping :smile:


No problem, anytime :smile:
(edited 6 years ago)
Because it should make a set between 0and 0.6 in the set 0 greater than x greater than 0.6 (SET A), where as greater than x greater than 0.6 (Set B)in this case if the set is integer it becomes a infinity set of all negative integers and positive till0.6 A={} if x€wA={}if x€iA={} if x€nB={0}if x€wB={.....,-1,0} if x€iB={} if x€n

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