If x + y + z > 0, is z > 1?
1. z > x + y + 1
2. x + y + 1 < 0
Please explain!
inequalities
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Given: x+y+z>0----equ 1
St 1: z>x+y+1
adding the two equations you get: x+y+2z>x+y+1
canceling x+y from both sides you get: 2z>1--->z>0.5-----INSUFF
St 2: x+y+1<0-----equ 2
we need to change the less than sign to greater than in order to add it with equ 1
Hence multiply -1 in equ 2 which gives us---> -(x+y+1)>0
Now add this equ with equ 1 and we get ----> z-1>0
ie z>1 SUFF
Hence B is the answer
St 1: z>x+y+1
adding the two equations you get: x+y+2z>x+y+1
canceling x+y from both sides you get: 2z>1--->z>0.5-----INSUFF
St 2: x+y+1<0-----equ 2
we need to change the less than sign to greater than in order to add it with equ 1
Hence multiply -1 in equ 2 which gives us---> -(x+y+1)>0
Now add this equ with equ 1 and we get ----> z-1>0
ie z>1 SUFF
Hence B is the answer