Question

How do you solve the inequality `1/(x+1)>3/(x-2)`?

Answer 1

`x < - 5/2 color(white)(xx)` or `color(white)(xx)-1 < x < 2`

Explanation:

First of all, note that your inequality is only defined if your denominators are not equal to zero:

`x + 1 != 0 <=> x != -1`

`x - 2 != 0 <=> x != 2`

Now, your next step would be to "get rid" of the fractions. This can be done if multiplying both sides of the inequality with `x+1` and `x-2`.

However, you need to be careful since if you multiply an inequality with a negative number, you must flip the inequality sign.

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Let's consider the different cases:

case 1: `color(white)(xxx) x > 2`:

Both `x + 1 > 0` and `x - 2 > 0` hold. Thus, you get:

`x - 2 > 3 (x + 1)`

`x - 2 > 3x + 3`

... compute `-3x` and `+2` on both sides...

`-2x > 5`

... divide by `-2` on both sides. As `-2` is a negative number, you must flip the inequality sign...

`x < - 5/2`

However, there is no `x` that satisfies both the condition `x > 2` and `x < - 5/2`. Thus, there is no solution in this case.

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case 2: `color(white)(xxx) -1 < x < 2`:

Here, `x + 1 > 0` but `x - 2 < 0`. Thus, you need to flip the inequality sign once and you get:

`color(white)(i) x - 2 < 3 (x + 1)`

`color(white)(x) -2x < 5`

... divide by `-2` and flip the inequality sign again...

`color(white)(xxx) x > -5/2`

The inequality `x > -5/2` is true for all `x` in the interval `-1 < x < 2`. Thus, in this case, we have the solution `-1 < x < 2`.

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case 3: `color(white)(xxx) x < -1`:

Here, both denominators are negative. Thus, if you multiply the inequality with both of them, you need to flip the inequality sign twice and you will get:

`x - 2 > 3x + 3`

`color(white)(i) -2x > 5`

`color(white)(xxi) x < - 5/2`

As the condition `x < -5/2` is more restrictive than the condition `x < -1`, the solution for this case is `x < - 5/2`.

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In total, the solution is

`x < - 5/2 color(white)(xx)` or `color(white)(xx)-1 < x < 2`

or, if you prefer a different notation,

`x in (- oo, -5/2) uu (-1, 2)`.

Answer 2

`]-oo, -5/2[ uu ]-1, 2[`

Explanation:

`1/(x+1)>3/(x-2)`

let pass everithing to the left side of the inequality by subtracting `3/(x-2)`:

`1/(x+1)-3/(x-2)>0`

Now we must, put all the inequation we the same denominator. The part with (x+1) we multiply by `(x-2)/(x-2)` (which is 1!) and vice-versa:

`(x-2)/((x+1)(x-2))-(3(x+1))/((x+1)(x-2))>0`

We did the trick before, to have all the inequation with the same denominator:

`(-2x-5)/((x+1)(x-2))>0`.

`(x+1)(x-2)` corresponds to a parabola which gives positive values in the ineterval `]-oo, -1 [ uu ] 2, +oo[` and negative values in the interval `]-1, 2[`. Remeber that x cannot be -1 or 2 due to giving denominator zero.

In the first case (denominator positive) we can simplify the inequation into:

`-2x-5>0` and `x in ]-oo, -1 [ uu ] 2, +oo[`

which gives:

`x<-5/2` and `x in ]-oo, -1 [ uu ] 2, +oo[`.

The interception of intervals above gives `x<-5/2`.

In the second case, the denominator is negative, so for the result giving a positive number, the numerator must be negative:

`-2x-5<0` and `x in ]-1, 2[`

which gives

`x>-5/2`. and `x in ]-1, 2[`

The interception of intervals gives `x in ]-1, 2[`

Joining the solutions of the two cases we obtain:

`]-oo, -5/2[ uu ]-1, 2[`