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Inequalities

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by parasite8 » Fri Sep 23, 2011 2:36 pm
Hi,

I have a question on inequalities with equations.

I tried using a strategy shown in Manhattan GMAT Prep - Guide 3 - Equations, Inequalities, & VICs on page 91 and 92.

When dividing the table states that divide and flip the extreme values. eg 8/LT2 = GT4 (if we know that LT2 is positive). Can I assume that if we do not know if LT2 is positive, then we can still divide and NOT flip or we just cannot divide to solve this question?

The reason I ask is because when I tried to solve: If b<2 and 2x-3b=0, which of the following must be true?
a. x>-3
b. x<2
c. x=3
d. x<3
e. x>3

I solved this by putting x to one side => x=(3b)/2
Then replaced "b" with LT2
x=(3LT2)/2
x=LT6/2
I wanted to divide here and x =GT3 (since I flipped the signs). My answer was incorrect and I want to know if I shouldn't have divided by 2 since I do not know if LT6 is positive or I should not have flipped the signs but can still divide.

(I ended up figuring it out a different way but I want to know why my original method did not work).

thanks,

Jen
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by pemdas » Fri Sep 23, 2011 7:21 pm
interesting method, never came acros such LT and GTs ...

i would follow 2x-3b=0 is equivalent to b=2x/3
and b<2 is equivalent to 2x/3<2 with resulting 2x<6 and x<3
d
parasite8 wrote:Hi,

I have a question on inequalities with equations.

I tried using a strategy shown in Manhattan GMAT Prep - Guide 3 - Equations, Inequalities, & VICs on page 91 and 92.

When dividing the table states that divide and flip the extreme values. eg 8/LT2 = GT4 (if we know that LT2 is positive). Can I assume that if we do not know if LT2 is positive, then we can still divide and NOT flip or we just cannot divide to solve this question?

The reason I ask is because when I tried to solve: If b<2 and 2x-3b=0, which of the following must be true?
a. x>-3
b. x<2
c. x=3
d. x<3
e. x>3

I solved this by putting x to one side => x=(3b)/2
Then replaced "b" with LT2
x=(3LT2)/2
x=LT6/2
I wanted to divide here and x =GT3 (since I flipped the signs). My answer was incorrect and I want to know if I shouldn't have divided by 2 since I do not know if LT6 is positive or I should not have flipped the signs but can still divide.

(I ended up figuring it out a different way but I want to know why my original method did not work).

thanks,

Jen
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by GMATGuruNY » Fri Sep 23, 2011 7:43 pm
parasite8 wrote:Hi,

I have a question on inequalities with equations.

I tried using a strategy shown in Manhattan GMAT Prep - Guide 3 - Equations, Inequalities, & VICs on page 91 and 92.

When dividing the table states that divide and flip the extreme values. eg 8/LT2 = GT4 (if we know that LT2 is positive). Can I assume that if we do not know if LT2 is positive, then we can still divide and NOT flip or we just cannot divide to solve this question?

The reason I ask is because when I tried to solve: If b<2 and 2x-3b=0, which of the following must be true?
a. x>-3
b. x<2
c. x=3
d. x<3
e. x>3

I solved this by putting x to one side => x=(3b)/2
Then replaced "b" with LT2
x=(3LT2)/2
x=LT6/2
I wanted to divide here and x =GT3 (since I flipped the signs)
No reason to flip the sign.
Since x = (less than 6)/2, x < 3.

I'm with pemdas.
Much easier to substitute the equation into the inequality than vice versa:
Since b<2, 3b < 6.
Since 2x-3b = 0, 2x = 3b.
Substituting 2x for 3b in the inequality:
2x < 6.
x < 3.
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by knight247 » Fri Sep 23, 2011 9:54 pm
@Jen
Ur method until x=LT6/2 is perfect. Now, you don't have to do any sign flipping here. In this case LT6/2 is LT3 and not GT3. Lets substitute values in here. Say 5.9/2 which equals 2.95 Hence the result is LT3. Its takes a little effort to master operations involving LT and GT. Try the following operations to strengthen ur fundamentals

8+LT2=
8-LT2=

8*LT2=
-7*LT2=
-7*LT(-7)=

8/LT2=
8/LT(-2)=

LT8*LT2=
LT(-8)*LT(-2)=
LT(-8)*LT(2)=
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by parasite8 » Sun Sep 25, 2011 1:21 pm
thanks for your responses. I'm not 100% comfortable to LT/GT method but Knight 247's practice questions definitely help me with understanding the concept better.
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