equalities

IMO answer is D
because:-

1. 2x - 3y = -2
since x^2 is always positive except when x is imaginary, so this is sufficient to answer.
2. x>2 and y>0
for all postive x >2 => x^2 > 2x
and for y>0
2x -3y < 2x < x^2
so this is also sufficient

Please share the original answer

beater wrote:Is 2x-3y<x^2?

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

Thanks!

IMO D.

x^2 will always be a positive, hence greater than -2. SUFFICIENT

Statement 2: 2x-3y < x^2 is same as -3y < x(x-2)

if y>0, the left side will always be negative and if x>2, right side will always be positive. SUFFICIENT

I understand the logic of the answers above me but how can we graphically answer this question.

I am drawing it (x^2 being a parabola with center O), and 2x-3y=2 being a line, and x^2 is below the line for certain values of x.

Please help!

beater wrote:Is 2x-3y<x^2?

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

Thanks!

this is how I understand

by using stmt 1. we are asked is -2<x^2 the answer is yes as x62 is always positive. SUFF

for 2 there can be two possibilities

x<y x>2 y>0

for putting integer values always, x^2 is greater.

or if x>y then minimum case x=2, y=0 where 2x-3y = x^2

for any other case when x>y and x>2 , y>0
2x-3y<x^2.
SUFF

hence D

IMO D

beater wrote:Is 2x-3y<x^2?

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

Thanks!

Statement 1. 2x-3y = -2.
assign this value in the inequality 2x-3y<x^2 it will becomes
-2 < x^2. No matter what ever the value of x, the x^2 will be positive. And this satisfy the inequality. Sufficient.

x>2 and y>0

if x = 3, y=1.... 3< 9 True.
if x = 3 and y = 1/3, 5<9 True.
if x=5/2 and y = 1/2, 3.5 < 6 (nearly) True
Sufficient.