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kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{2x3}{x+1}<1\]
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
I would split it into two cases
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
how would you do that
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I would first write as f(x)<0 Then find where f(x) is zero and undefined Then draw a number line with the numbers that make f zero or undefined Then test the intervals around those numbers to see if f is less than 0 or not
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Since we are to multiply (x1) on both sides, I would split it into x1<0 and x1>0
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
you mean x+1?
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Yes sorry
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
SO i take x+ 1 and get \[2x3<x+1 and 2x3>x+1\]
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Well actually I believe we should do it case by case
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
When x+1<0: 2x3>x+1 ... When x+1>0: 2x3<x+1 ...
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
so then my next step would to add in the 3 on both sides corect? so it would turn into \[2x>x+4\] and \[2x<x+4\]
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
And for presentation purpose I would write it as 2x>x+4 where x+1<0 and 2x<x+4 where x+1>0
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
then do I divide by 2 on both of the equations or divide by 2x?
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
You subtract x from both sides
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
so It would be \[x2x>4 where x+1<0\] and just the oposit sign on the otherone?
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
No, it should be 2xx>4 where x+1<0
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
ok now im starting to understand so then the next step would be to hmove the 4 over
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Nah, the next step is subtracting it
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
ok so it would then be 2xx4> 0 and where x+1<0 and 2xx4< 0 and where x+1>0 and t
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
I mean you can write 2xx as x
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
When x+1<0: 2x3>x+1 2x>x+4 2xx>4 x>4 When x+1>0: 2x3<x+1 2x<x+4 2xx<4 x<4
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
However, when x+1<0, x can't be >4, and the same goes to the second case
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Therefore there is no solution
 one month ago

aum Group TitleBest ResponseYou've already chosen the best response.1
The inequality is valid for x in the interval (1, 4).
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
@aum tell me what is wrong in my solution, I have no time to doublecheck
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
@aum I thought so. I see how to get the 4 but not the 1 could you help me solve this?
 one month ago

aum Group TitleBest ResponseYou've already chosen the best response.1
My method is what @myininaya suggested. (2x  3) / (x + 1) < 1 (2x  3) / (x + 1)  1 < 0 (2x3  x  1) / (x + 1) < 0 (x  4) / (x + 1) < 0 The x values that are of interest to us here are the ones that will make f(x) = (x4)/(x+1) zero or undefined. f(x) = 0 when x = 4 f(x) is undefined when x = 1. So the number line is split into three intervals: (infinity, 1); (1, 4), (4, infinity). Pick a convenient number in each interval and see if the inequality is valid. f(x) = (x  4) / (x + 1) < 0 when x = 2, f(x) = (6) / (1) = 6 which is greater than 0. So not a solution. when x = 0, f(x) = 4/1 = 4 which is less than 0. This is a solution. when x = 5, f(x) = 1 / 6 which is greater than 0. Not a solution. So the solution is x in the interval (1, 4).
 one month ago

calebcalebhoffman Group TitleBest ResponseYou've already chosen the best response.0
thank you so much
 one month ago

aum Group TitleBest ResponseYou've already chosen the best response.1
you are welcome.
 one month ago

aum Group TitleBest ResponseYou've already chosen the best response.1
To follow @kc_kennylau 's method: (2x  3) / (x + 1) < 1 Case 1: x + 1 > 0 Since (x+1) is positive, we can multiply both sides by (x+1) and the "less than" inequality will remain intact. 2x  3 < x + 1 x < 4 Case 2: x + 1 < 0 Since (x+1) is negative, multiplying both sides by (x+1) would change the "less than" inequality to a "greater than" inequality. 2x  3 > x + 1 x > 4. But for the second case we assumed x + 1 < 0 which implies x < 1 But we get the solution as x > 4. x cannot be less than 1 and also be greater than 4 simultaneously. Thus, case 2 yields NO solution. Therefore, there is no solution in the domain x < 1. Combining the two cases we get x should be less than 4 but greater than 1.
 one month ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
I incorrectly rejected case 1
 one month ago
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