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calebcalebhoffman
 11 months ago
how would I solve for x if 2x3/x+1<1
calebcalebhoffman
 11 months ago
how would I solve for x if 2x3/x+1<1

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

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

calebcalebhoffman
 11 months ago
Best ResponseYou've already chosen the best response.0how would you do that

myininaya
 11 months ago
Best ResponseYou've already chosen the best response.1I 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

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

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

kc_kennylau
 11 months ago
Best ResponseYou've already chosen the best response.0Well actually I believe we should do it case by case

kc_kennylau
 11 months ago
Best ResponseYou've already chosen the best response.0When x+1<0: 2x3>x+1 ... When x+1>0: 2x3<x+1 ...

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

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

calebcalebhoffman
 11 months ago
Best ResponseYou've already chosen the best response.0then do I divide by 2 on both of the equations or divide by 2x?

kc_kennylau
 11 months ago
Best ResponseYou've already chosen the best response.0You subtract x from both sides

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

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

calebcalebhoffman
 11 months ago
Best ResponseYou've already chosen the best response.0ok now im starting to understand so then the next step would be to hmove the 4 over

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

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

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

kc_kennylau
 11 months ago
Best ResponseYou've already chosen the best response.0When 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

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

kc_kennylau
 11 months ago
Best ResponseYou've already chosen the best response.0Therefore there is no solution

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

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

calebcalebhoffman
 11 months ago
Best 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?

aum
 11 months ago
Best ResponseYou've already chosen the best response.1My 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).

calebcalebhoffman
 11 months ago
Best ResponseYou've already chosen the best response.0thank you so much

aum
 11 months ago
Best ResponseYou've already chosen the best response.1To 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.

kc_kennylau
 11 months ago
Best ResponseYou've already chosen the best response.0I incorrectly rejected case 1
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