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

##### 2 Answers

#### Answer:

#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

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 - 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

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

Here,

#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

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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

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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:

#### Explanation:

let pass everithing to the left side of the inequality by subtracting

Now we must, put all the inequation we the same denominator. The part with (x+1) we multiply by

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

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

which gives:

The interception of intervals above gives

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

which gives

The interception of intervals gives

Joining the solutions of the two cases we obtain: