LulaBrazilia wrote:Is 2x - 3y < x²?

1) 2x - 3y = -2
2) x > 2 and y > 0

Target question: Is 2x - 3y < x²?

Statement 1: 2x - 3y = -2
We can now take the target question, and replace 2x - 3y with -2 to get: Is -2 < x²?
Since x² must be greater than or equal to zero (as with all SQUARED numbers), we can conclude that it MUST be the case that -2 < x²
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x > 2 and y > 0
This one is a little trickier.
First recognize that if y > 0, then 3y = some positive value. So, if we take 2x and SUBTRACT 3y, we get a lesser value.
In other words, we can be certain that 2x - 3y < 2x

Now let's compare 2x and x²
If x > 2 (as we're told in statement 2), x² MUST be greater than 2x (test it out if you're not convinced)
In other words, it MUST be the case that 2x < x²
Now that we're derived two inequalities, we can combine them to get 2x - 3y < 2x < x²
From this, we can see that it MUST be the case that 2x - 3y < x²
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent