Inequality Question.

\((x+1)*(\frac{1}{|2-x|}+\frac{1}{x-5}) <= 0\)

For this to be true, either the first term is positive and the second negative or vice versa

Case 1 : x+1 >=0 ... x>=-1

\(\frac{1}{|2-x|}+\frac{1}{x-5} <= 0\)
\(\frac{1}{|2-x|}<=\frac{1}{5-x}\)

Note that since |2-x| is always positive, to have 1/(5-x) greater than 1/(|2-x|), it also has to be always positive. which means x<5.

If 2-x>=0 OR x<=2, then this implies 5-x<=2-x , which is impossible

So 2-x<=0 OR x>=0, in which case we get :
5-x<=-(2-x) OR x>=7/2


So we get the range [7/2,5)

Case 2 : x+1 <=0 OR x<=-1

Hence \(\frac{1}{|2-x|}>=\frac{1}{5-x}\)
Since |2-x| is always positive all we need to ensure is that 5-x is a smaller number than |2-x|

5-x >= |2-x|

If 2-x >=0 , i.e, x<=2
then all values of x satisfy
5-x >= 2-x OR 5>=2

If 2-x <0 , its impossible since we already know x<=-1

Hence all numbers in range satisfy (-infinity, -1]


Combining the answer is :

(-infinity,-1] U [7/2,5)

Hi shrouded1

Could you please explain this bit :

If 2-x>=0 OR x<=2, then this implies 5-x<=2-x

Regards,
Subhash

I am trying to solve an inequality in which one of the sides is an absolute value. In such a case, you have to assume two cases, first that the expression inside the absolute value is positive and then that it is negative. So this is simply the first of those two cases.

After taking the case, we remove the absolute value and try to solve the inequality (which in this case, is converted from an 1/Expression form to an Expression form easily since we know both sides are now positive)

Hi Could you please explain
If 2-x>=0 OR x<=2, then this implies 5-x<=2-x

Thanks

Thank you so much, Shrouded 1.

However could anyone tell me whether there are any shortcuts for this inequality?

It takes me too long to solve it. I would like to know how is it possible to solve this inequality within 2 minutes.

Steve.

Also,

I would like to know what types of inequalities are there on the test.

Thank you,
Steven.

good question indeed.

Thank you,
Amit.

Would anyone be able to elaborate on it?

Thank you so much,
Steve.