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

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

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

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

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

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

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Well actually I believe we should do it case by case
 2 months 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 ...
 2 months 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 \[2x<x+4\]
 2 months 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
 2 months 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?
 2 months ago

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

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

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

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

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

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

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

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

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