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(x/(x-2))-1 > 0
Created 24th November 2013 @ 15:30
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How do I solve this? I’ve tried every possible way so far yet I get the wrong answer
whats the point
x>2
Quoted from atmo
x>2
yes, I know but how?
(x/(x-2))-1 > 0
Why can’t I just move the 1 to the right side?
Quoted from TWEEKARN\
whats the point
what do you think…
x/x-2 – 1 > 0
x/x-2 – x-2/x-2 > 0
x-x+2/x-2 > 0
2/x-2 > 0
2*(x-2) > 0
2x – 4 > 0
2x > 4
x > 2
Last edited by Bonkers,
(x/(x-2)) – 1 > 0
x(x-2) – 1 (x-2)^2 > 0
x^2 – 2x – (x^2 -4x + 4) > 0
x^2 – 2x – x^2 + 4x – 4 > 0
2x – 4 > 0
x > 2
if you multiply the whole line it has to be with a positive number so that the inequality does not change sign, so you use the square of the denominator (x-2).
Quoted from Bonkers
x/x-2 – 1 > 0
x/x-2 – x-2/x-2 > 0
x-x+2/x-2 > 0
2/x-2 > 0
2*(x-2) > 0
2x – 4 > 0
2x > 4
x > 2
2/x-2
2*(x-2)
Lines 4,5 wtf
Quoted from atmo
(x/(x-2)) – 1 > 0
x(x-2) – 1 (x-2)^2 > 0
x^2 – 2x – (x^2 -4x + 4) > 0
x^2 – 2x – x^2 + 4x – 4 > 0
2x – 4 > 0
x > 2
thanks, why can’t I multiple by x-2 at the beginning though
Last edited by jx53,
Quoted from Bonkers
if a/b > 0 then a*b > 0
so (a/b > 0) = (a*b > 0)?
Why can’t I then?
(X/(X-2))-1 > 0
x^2-2x-1 > 0
that gives wrong answer :I
Last edited by jx53,
Quoted from jx53
Quoted from atmo
(x/(x-2)) – 1 > 0
x(x-2) – 1 (x-2)^2 > 0
x^2 – 2x – (x^2 -4x + 4) > 0
x^2 – 2x – x^2 + 4x – 4 > 0
2x – 4 > 0
x > 2thanks, why can’t I multiple by x-2 at the beginning though
Because you don’t know what value x-2 has at this point. If x turned out to be 0.5, then x-2 would be a negative number so you would be multiplying your inequality be a negative which means you have to change the inequality sign from > to <. That's why you square it to ensure that you are multiplying by a positive.
Quoted from atmo
[…]
Because you don’t know what value x-2 has at this point. If x turned out to be 0.5, then x-2 would be a negative number so you would be multiplying your inequality be a negative which means you have to change the inequality sign from > to <. That's why you square it to ensure that you are multiplying by a positive.
That sounds somewhat smart. Thanks
Quoted from jx53
[…]
what do you think…
that youll never have use for this at all
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