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Re: Real numbers x, y, and z satisfy the inequalities [#permalink]
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Frankly I do not know why he did not consider x

However, if you pick numbers you can reach the same solution
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Re: Real numbers x, y, and z satisfy the inequalities [#permalink]
How so Carcass? because if we add we'll get 0<x+y+z<3 = x+y+z but we don't have that in the option
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Real numbers x, y, and z satisfy the inequalities [#permalink]
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I do not know how you get , summing the inequality, the three

E is the only option that is positive. Even because y is negative and z is positive but certainly z > y
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Re: Real numbers x, y, and z satisfy the inequalities [#permalink]
Pls can you solve the question by steps? I don't seem to understand
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Re: Real numbers x, y, and z satisfy the inequalities [#permalink]
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x is positive and a fraction such as 1/2

y is negative and a fraction such as -1/2

z is a positive fraction such as 1.5 or 3/2

All the answer choices but E are negative or positive it depends on their value

In E even if you take a value for y really close to -1 and in z a value really close to 1 I.E a minimum and a maximum value the result will be always positive
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