Inequalities

[juan_ignacio14]

Hi, I found two DS exercises of inequelities in the CAT examen and couldn't find them here.
Here they are:

1) If x is an integer, what is the value of x? (Ans: C)
(1) -5x > -3x + 10
(2) -11x - 10 < 67

2) If x is an integer, what is the value of x? (Ans: C)
(1) 2x - 3 < 5
(2) -4x < -8

They are simple and I was able to resolve them BUT I don't understand why they don't take into account if X is negative. In the explanation they only resolve for possitive X. Example for 1) (1)
-5x > -3x + 10
5x - 3x < -10 (don't forget: switch the sign when multiplying or dividing by a negative)
2x < -10
x < -5

That was the CAT's explanation. When I did the exercise I also resolved for negative X

-5x > -3x + 10
-2X > 10 (if x is negative...)
x> 5

What am I doing wrong?

[LazyNK]

Hi Juan,
The method you are using - deciding first whether x is negative or positive, and building the inequality accordingly - is not required for this case.
You need to create the equation/inequation specifically considering the two cases of positive and negative variable ONLY in case the variable is inside an absolute value symbol.
-NK

[juan_ignacio14]

Hi. Thank you. Could you please give me the reason of this and if possible five me an example on when should I evaluate for neg and pos X? What about squares or roots?
Thanks

[LazyNK]

1.) Example of Absolute Value where you need to consider both positive and negative value of x :
|x| > 4
-> if x is positive, x>4
-> if x is negative -x > 4 -> x < -4

2.) For squares and roots, you need not take the two cases separately. Simply arrive at the final equation in terms of x2 or sauareroot(x), and solve -both positive and negative values for x2 . eg. x2=4 -> x=+/-2, squareroot(x)=2 -> x=4, squareroot(x)=-2 ->x=4
-NK

[tim]

Thanks Lazy. Juan, let us know if you have any other questions..