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Mathematics 14 Online
OpenStudy (anonymous):

Choose A :- If the question can be answered by both the statement together but cannot be answered by using either statement alone. Choose B :- If the question can be answerd using any statement alone. Is X-Y positive? 1. Four times X is 2 more than Y 2. X-2Y is positive

OpenStudy (anonymous):

must be B since first statement says \[4x=y+2\] you cannot conclude from this that \[x-y>0\]

OpenStudy (anonymous):

whoa big typo above sorry

OpenStudy (anonymous):

let me try again. knowing that \[4x=y+2\] does not prove \[x-y>0\] we can can solve \[x-(4x+2)>0\] \[-3x-2>0\] \[-3x>2\] \[x<-\frac{2}{3}\] telling us that IF \[4x=y+2\] THEN \[x-y>0\] only if \[x<-\frac{2}{3}\]

OpenStudy (anonymous):

I can tell you the answer of the question is with the help of both statement only we can conclude that X-Y positive. I am not able to find the explaination for the same

OpenStudy (anonymous):

Neat question for mental gyrations:-) Take the second condition and write as (x-y) -y so is it possible for x-y to be positive and the whole expression negative or 0? If x is 1 and y is 1/2, yes it is, so knowing that x-2y is not positive does not help u determine that x-y is positive. I'm trying to find a way to explain it in English...

OpenStudy (anonymous):

i think after i fixed by typo my answer is correct. we know that if the first statement is true, then x - y is positive only if \[x<-\frac{2}{3}\]

OpenStudy (anonymous):

It require both statement to coclude the answer

OpenStudy (anonymous):

of course it is possible that \[x>-\frac{2}{3}\] which would make \[x-y>0\] a false statment

OpenStudy (anonymous):

Well if satellite is right in the first answer and I am right in the second, then u need both...

OpenStudy (anonymous):

x−(4x+2)>0

OpenStudy (anonymous):

is this correct?

OpenStudy (anonymous):

as 4x = y+2

OpenStudy (anonymous):

Can't we say statement 2 is enough to answer the question.There is no need of statement 1

OpenStudy (anonymous):

maybe. lets check

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

second statement not enough. reason one would be to graph the region \[x-2y>0\] and see that the region \[x-y>0\] is not contained in it. second and easier reason is just pick an example. put \[x=-4,y=-3\] then \[x-2y=-4-2\times -3=-4+6=2>0\] so that statement it true. but \[x-y=-4+3=-1<0\] so first statement would be false.

OpenStudy (anonymous):

oh estudier sorry you got this already!

OpenStudy (anonymous):

The thing that the two statements together gives u is a lower numerical bound whereas separately you only have one variable in terms of the other.

OpenStudy (anonymous):

your example was right i just wasn't paying enough attention. so the answer is you need both

OpenStudy (anonymous):

The lower numerical bound prevents the construction of counterexamples.

OpenStudy (anonymous):

Good team work..!Appreciate it guys.

OpenStudy (anonymous):

Nice question...

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