GraphicallyOne easy way to do this is to see then the lines intersect by graphing them. In the attached, red is (1), blue is (2), and yellow is the prompt. You can see that (1) intersects with y>x when x, y negative, and thus you can pick options that allow the red line to be true, but can literally go on either side of y > x. For the blue line, that is also possible, but only when y and x are positive. Thus, we know that (2) is sufficient for all x,y negative.

AlgebraicallyAnother easy way to do this is to see when the lines intersect, but by using algebra.

(1) 2y > 3x+1

Pretend this is 2y=3x+1.

Set this equal to what we are comparing it to, y=x, or, 2y=2x.

2y=3x+1=2x

They intersect at x = -1. Either side of this, you will a Y or N answer when seeing if (1) is sufficient.

(2) 3y > 2x+1

Pretend this is 3y=2x+1.

Set this equal to what we are comparing it to, y=x, or, 3y=3x.

3y=2x+1=3x

They intersect at x=1. If less than 1, you will always get the same answer when seeing if (2) is sufficient. Since x < 0 < 1, (2) is sufficient.

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