anonymous
  • anonymous
how would I solve for x if 2x-3/x+1<1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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kc_kennylau
  • kc_kennylau
\[\frac{2x-3}{x+1}<1\]
kc_kennylau
  • kc_kennylau
I would split it into two cases
anonymous
  • anonymous
how would you do that

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myininaya
  • myininaya
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
kc_kennylau
  • kc_kennylau
Since we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0
myininaya
  • myininaya
you mean x+1?
kc_kennylau
  • kc_kennylau
Yes sorry
anonymous
  • anonymous
SO i take x+ 1 and get \[2x-3x+1\]
kc_kennylau
  • kc_kennylau
Yes
kc_kennylau
  • kc_kennylau
Well actually I believe we should do it case by case
kc_kennylau
  • kc_kennylau
When x+1<0: 2x-3>x+1 ... When x+1>0: 2x-3
anonymous
  • anonymous
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
kc_kennylau
  • kc_kennylau
Yep
kc_kennylau
  • kc_kennylau
And for presentation purpose I would write it as 2x>x+4 where x+1<0 and 2x0
anonymous
  • anonymous
then do I divide by 2 on both of the equations or divide by 2x?
kc_kennylau
  • kc_kennylau
You subtract x from both sides
anonymous
  • anonymous
so It would be \[x-2x>4 where x+1<0\] and just the oposit sign on the otherone?
kc_kennylau
  • kc_kennylau
No, it should be 2x-x>4 where x+1<0
anonymous
  • anonymous
ok now im starting to understand so then the next step would be to hmove the 4 over
kc_kennylau
  • kc_kennylau
Nah, the next step is subtracting it
anonymous
  • anonymous
ok so it would then be 2x-x-4> 0 and where x+1<0 and 2x-x-4< 0 and where x+1>0 and t
kc_kennylau
  • kc_kennylau
I mean you can write 2x-x as x
kc_kennylau
  • kc_kennylau
When x+1<0: 2x-3>x+1 2x>x+4 2x-x>4 x>4 When x+1>0: 2x-3
kc_kennylau
  • kc_kennylau
However, when x+1<0, x can't be >4, and the same goes to the second case
kc_kennylau
  • kc_kennylau
Therefore there is no solution
kc_kennylau
  • kc_kennylau
Wait
aum
  • aum
The inequality is valid for x in the interval (-1, 4).
kc_kennylau
  • kc_kennylau
@aum tell me what is wrong in my solution, I have no time to double-check
kc_kennylau
  • kc_kennylau
bye
anonymous
  • anonymous
@aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?
aum
  • aum
My method is what @myininaya suggested. (2x - 3) / (x + 1) < 1 (2x - 3) / (x + 1) - 1 < 0 (2x-3 - 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) = (x-4)/(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).
anonymous
  • anonymous
thank you so much
aum
  • aum
you are welcome.
aum
  • aum
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.
kc_kennylau
  • kc_kennylau
I incorrectly rejected case 1
aum
  • aum
yes.

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