eddyed911
If x+y+z > 0, is z > 1 ?
1) z > x + y + 1
2) x + y + 1 < 0
this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough.
Thanks,
Eddy
In case you get confused with addition/subtraction of inequalities, you can stick with addition only, if you like. Just rewrite one inequality to give them the same sign. Remember that you can add inequalities only when they have the same sign.
Ques: Is z > 1?
Given: x+y+z > 0 .......... (I)
Stmnt 1: z > x + y + 1 .........(II)
(I) and (II) both have the same inequality sign '>' so we can add them.
x + y + 2z > x + y + 1
We get, z > (1/2)
z may or may not be greater than 1. Not sufficient.
Stmnt 2: x + y + 1 < 0
We can re-write this as 0 > x + y + 1 .......(III)
Now, (I) and (III) have the same sign '>' so you can add them.
x + y + z > x + y + 1
We get z > 1. Sufficient.
Answer (B)
To clarify on Bunuel's explanation at the start of this form, "You can only apply subtraction when their signs are in the opposite directions:
If a>ba>b and c<dc<d (signs in opposite direction: >> and <<) --> a−c>b−da−c>b−d (take the sign of the inequality you subtract from).
Example: 3<43<4 and 5>15>1 --> 3−5<4−13−5<4−1 ." I thought that you were not allowed to subtract nor divide inequalities? Why does Bunuel show how to subtract inequalities here?