Is 2x-3Y<x^2
1) 2x-3y=-2
2) x>2 and Y>0

Can you help me understand how I approach the 2nd statement?
Thanks,
Mariela :shock:

Mariela,

This is how I would solve the question:

(I) Is -2 < x^2 ? Yes, because anything to the power of 2 is going to be either 0 or positive. Therefore, sufficient.

(II) I would test out a couple of numbers here.

Test 1:
X = 2.1 and Y = 0.1

2(2.1) - 0.3 < (2.1)^2 True

Test 2:
X = 2.1 and Y = 100

2(2.1) - 300 < (2.1)^2 True

Test 3:
X = 100 and Y = 0.1

200 - 0.3 < 100^2 True

Test 4:
X = 100 and Y = 100

200 - 300 < 100^2 True

I am pretty sure there are better ways to solve, but this is how I would approach it and I would select answer D. It took less than 2 minutes. :)

Hope that helps!

Is 2x-3Y<x^2
1) 2x-3y=-2
2) x>2 and Y>0

Mariela -

To answer whether (2) is sufficient by itself, it helps to rearrange the original inequality (rephase the question):

2x-3y<x^2
2x-x^2<3y
x(2-x)<3y

Given the inequality above consider (2): If x>2, then the left side of the inequality is always negative, because the x is positive and the quantity (2-x) will be negative. If y>0 then the quantity 3y is always positive, so it will always be the case that 2x-3y<x^2 and (2) by itself is sufficient to answer the question.

Plugging in numbers like the previous poster did is a perfectly valid way to approach these problems - but you have to be v. careful not to miss a region of interest when choosing numbers to test. If you become comfortable and quick with algebra, it can often give you a definitive answer quickly. One important caveat: Be v. careful manipulating inequalities that contain variables that might or might not be negative. In the example above the inequality was manipulated using addition and subtraction only - if you divide or multiply by a variable or quantity that might or might not be negative, you have to account for both cases and remember to flip the inequality sign accordingly.

Best,
Jeff[/i][/b]

Beautiful, Jeff!

Yes, using a plug and play method is risky as you may miss a region (as Jeff stated).

Mariela, I would suggest trying to solve something algebraically first, but do not spend a lot of time if you get nowhere or cannot "see" the algebraic solution; if you get stuck, fall back to the plug and play.

Good solutions, and excellent point about the potential pitfalls of plug-and-play.

Although hard to reproduce here, you may want to try this on your own: graphing.

The question rephrases to "Is y > (2/3)x - x^2?" Plot (2/3)x - x^2 and you get an upside-down "U" centered just to the right of the y-axis, with the base of the "U" just above the x-axis. The visual interpretation of the rephrase is "Are all of the points (x, y) in the region above the 'U'?"

(1) Rephrases to the equation of a line y = (2/3)x + 2/3, which crosses the y-axis at 2/3 (above the "U") and has a slope of 2/3. The entire line is above the "U." SUFFICIENT.

(2) The rectangularly bounded region to the right of x = 2 and above y = 0. This entire region is above the "U," too. SUFFICIENT.

Obviously, this method is not required for this problem, and there is a little set-up time involved in drawing the "U." Such a method could be a life-saver on a problems with more complicated constraints, or when you want to plug-and-play but have no idea what values to try. Also, some people are more visually inclined and can follow this approach better than the algebraic one.