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

2 Answers
Jan 19, 2016

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 #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.

=========================================

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.

=========================================

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#.

=========================================

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#.

=========================================

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)#.

Jan 26, 2016

Answer:

#]-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[#