Bunuel wrote:
Is \(x > 1\)?
(1) \(x^2 + x + 2 > 8\)
(2) \(8(x – 4) > 4(x – 2)\)
Instead of solving explicitly, we'll look for simple numbers that contradict the given statements.
This is an Alternative approach.
(1)
Let's try a number that gives us a YES. If we make x very large, say x = 100, then 100^2+100+2 is definitely larger than 8.
Not let's try to contradict this by getting a NO. If we make x very small, say x = -100 then (-100)^2 - 100 + 2 is 10,000 -100 + 2 which is also larger than 8.
We have found both x > 1 and x < 1 which gives a correct statement (1) so this is not enough.
Insufficient!
(2)
Since it's hard to guess what numbers to pick, we'll first simplify a bit.
8x - 32 > 4x - 8
4x > 24
x > 6.
In this case, there is no need to pick numbers at all!
Sufficient.
(B) is our answer.
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