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Durgesh or gmatnub - can you explain this a bit more? I understand you've graphed out the inequalities. One inequalty create the yellow shaded area and the other inequality creates the green shaded area. Can you go into more detail regarding their relationship with each other? _________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Durgesh or gmatnub - can you explain this a bit more? I understand you've graphed out the inequalities. One inequalty create the yellow shaded area and the other inequality creates the green shaded area. Can you go into more detail regarding their relationship with each other?
its just visulizing the inequalities with two variables....
for example if you have only one variable ... one of the ways of doing such problems is draw anumber line and mar the portion of the number line which falls in that rang
what is the value of x, an integer 1) 2 < x < 8 2) 6 < x < 10
draw a number line mark the segament between 2 and 8 mark the segament between 6 and 10
the common segamant will give the values of x, which will satisfy both.... as x is an integer, th only possible value is 7
Going back to our question, the idea is to find a target area which is represented by one side of x+y=0
by combinng the two conditions, we can find an area which will always be on one side if x+y=0, no matter what is the value of x and y, this Suff.
so what is the significance of the part on the top right that does not overlap at all?
In this question no significance actually, that area represents values of x,y which we dont have to consider..... if you are trying to solve the question by plugging values, no value in that area will satisfy (1) or (2)....
I wouldn't recommend graphic approach for this problem, algebraic approach is simpler and fairly straightforward:
Is m+z > 0?
(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.
(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.
Re: is m+z>0? 1.m-3z>0 2.4z-m>0. is this diag ok,? if [#permalink]
27 Apr 2014, 18:32
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