how would I solve for x if 2x-3/x+1<1

- anonymous

how would I solve for x if 2x-3/x+1<1

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- kc_kennylau

\[\frac{2x-3}{x+1}<1\]

- kc_kennylau

I would split it into two cases

- anonymous

how would you do that

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## More answers

- 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

Since we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0

- myininaya

you mean x+1?

- kc_kennylau

Yes sorry

- anonymous

SO i take x+ 1 and get \[2x-3x+1\]

- kc_kennylau

Yes

- kc_kennylau

Well actually I believe we should do it case by case

- kc_kennylau

When x+1<0:
2x-3>x+1
...
When x+1>0:
2x-3

- 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

Yep

- kc_kennylau

And for presentation purpose I would write it as
2x>x+4 where x+1<0
and
2x0

- anonymous

then do I divide by 2 on both of the equations or divide by 2x?

- kc_kennylau

You subtract x from both sides

- anonymous

so It would be \[x-2x>4 where x+1<0\] and just the oposit sign on the otherone?

- kc_kennylau

No, it should be 2x-x>4 where x+1<0

- anonymous

ok now im starting to understand so then the next step would be to hmove the 4 over

- kc_kennylau

Nah, the next step is subtracting it

- 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

I mean you can write 2x-x as x

- 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

However, when x+1<0, x can't be >4, and the same goes to the second case

- kc_kennylau

Therefore there is no solution

- kc_kennylau

Wait

- aum

The inequality is valid for x in the interval (-1, 4).

- kc_kennylau

@aum tell me what is wrong in my solution, I have no time to double-check

- kc_kennylau

bye

- anonymous

@aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?

- 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

thank you so much

- aum

you are welcome.

- 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

I incorrectly rejected case 1

- aum

yes.

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