Lesson: Inequalities - Part I

Comment on Inequalities - Part I

I solved it in another way and the result is slightly different.

5(1-2x)+7x > 4-x+9+2x

5-3x > 13+x
-2x > 8
x < -4

Is it right also? Why does this difference occur?
greenlight-admin's picture

I'm not sure what "it" refers to (as in "I solved IT in another way and the result is slightly different.") Are you referring to a specific practice question or example?

That said, the third step in your solution is incorrect.

Take: 5 - 3x > 13 + x
Subtract x from both sides to get: 5 - 4x > 13
Subtract 5 from both sides to get: -4x > 8
Divide both sides by -4 to get: x < -2

Cheers,
Brent

Hello Brent,

For the following question, when i try to solve by isolating variables to one side, I end up having 1 (in Qty A) and 2y-x in (QTY B). 2y can be maximum when y=1/2. As 1-x will result in a value less than 1, I got result as A.

However, the response given by Sandy also seems valid (That the solution is D). Am I missing anything in my calculation ? Can you please help me ?

http://greprepclub.com/forum/given-that-0-x-y-1806.html

Thanks :)
greenlight-admin's picture

Question link: http://greprepclub.com/forum/given-that-0-x-y-1806.html

Your statement that "2y can be maximum when y = 1/2" is incorrect.

All we know about y is that it's greater than both 0 and x, AND it's less than 1. So, y could be something like 0.9 or even 0.99999

Got my mistake. Thanks (y)

http://greprepclub.com/forum/x-is-the-set-of-all-integers-n-that-satisfy-1808.html
Brent, i am so confused with this one... please help me out...
2 ≤ |n| ≤ 5 how is n getting negative values?
greenlight-admin's picture

Hello Brent!
I had a doubt in this reinforcement question

If x, y, and z are positive numbers such that 3x < 2y < 4z, which of the following statements could be true?

Indicate all such statements.

A) x = y

B) y = z

C) y > z

D) x > z


For this question is it the case that if u find any such possible values for x y or z then u can just go ahead and mark the answer correct.
and what if u find a value which dosen't hold true for eg in choice D if we take x=4 and z=3 then we get that the inequality is equal and not 3x<4z .Still choice D is marked as a correct answer
I had this confusion .please help!
greenlight-admin's picture

This comes down to MUST be true versus COULD be true.

If the question asks us "What COULD be true?", then we need only find one instance where that thing is true. If we find an instance where a statement is not true, we can't eliminate that answer, since it doesn't rule out the possibility that the statement COULD be true.

If the question asks us "What MUST be true?", then we can ELIMINATE a a statement if it turns out to be false. However, we must be certain that the statement is ALWAYS true before we can select it.

----------------------------------------------

That part was more of an ASIDE, since your solution does not really apply to the above concepts.

In your solution, you are REVERSING the question.
The question says IF 3x < 2y < 4z, THEN what could be true.
So, for example, IF 3x < 2y < 4z, THEN could it be true that x > z (answer choice D).

In your solution, you are asking "IF x > z, THEN could it be true that 3x < 2y < 4z?

So, in order to check each of the 4 statements, you must first find values of x, y and z that satisfy the given inequality (3x < 2y < 4z).

Does that help?

Cheers,
Brent

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