Mod Query Important

[CrackGMAC]

I have two general queries in Modlus questions

1. When we say X < Y does that mean x-y < 0 or
x/y < 1. In my view both are same. If not please let me know what the difference is and where we follow these different rules.
2. What is best approach to solve the questions when we have mod on both the sides,

say for example x^2 + |x+2| < x^4 + |x-10

[jd006h]

if x<y, then x - y <0. ALWAYS.

if x<y, then x/y < 1, ONLY IF y is positive. when you divide both sides of an inequality by a negative, it switches direction of the inequality sign. this makes sense: -3<-1, but when you divide out, 3 is not less than 1.

[heshamelaziry]

I had the same confusion and made this rule:

DO not divide, multiply, or cross multiply variables on opposite sides of an inequality by the othe variable, unless you know the sign of both variables and they are the same.

If you have a problem and says x > y, it is better to leave it as its. My experience that it is easier to work with that x-y >0

There was a question or two here that I did this with and helped get a more clear answer with less confusion.

Also, if you have this: 1/3 > 1/x > 1/2 write as follows 3 < x < 2

[palvarez]

I have two general queries in Modlus questions

1. When we say X < Y does that mean x-y < 0 or
x/y < 1. In my view both are same. If not please let me know what the difference is and where we follow these different rules.
2. What is best approach to solve the questions when we have mod on both the sides,

say for example x^2 + |x+2| < x^4 + |x-10

x < y
x + a < y + a
x - a < y -a
x/k < y/k when k is +ve
x/k > y/k when k is negative

All the above same.

x < y
subtract y from both sides, the inequality doesnt change.

so, x -y < 0

x/y < 1

here, we don't know whether y is posiitve or not.

When y is positive, x < y
when y is negative, x > y.

My approach here is to "not bring verbal reasoning": just manipulate algebraically.

x/y < 1
xy/y^2 < 1
xy - y^2 < 0 (we can shuffle around y^2, since it is +ve)

y(x-y) < 0

The above inequality tells us two cases.

(1) y is +ve and x-y is +ve
(2) y is -ve and x-y is -ve.

At this stage, you can translate algebra into words.



x^2 + |x+2| < x^4 + |x-10|

In questions like this, there is a default method.

we got two critical points to look at: -2 and 10

So, solve for each range.

(-inf, -2): x^2 -x -2 < x^4-x +10
x^4 -x^2 + 12 > 0
True for all x's in this range
(-2, 10): x^2 + x+2 < x^4 -x + 10
x^4 -2x^2 +8 > 0
True for all x's in this range
(10, +inf): x^2 + x +2 < x^4 + x -10
x^4 -x^2 -12 > 0
(x^2 - 4)(x^2+3) > 0
x^2 - 4 > 0
or true for all x's.

You can also use conceptual approach to get rid of odd choices