The line passes through (6,0) and (0,3)

Slope= 3/-6 = -1/2

y = (-1/2) x + b

y intercept b is the value at which x=0

3= 0 +b

eq of line is y = ( -1/2 ) x + 3

Q is asking for an inequality for all the points beneath (read beneath less than)

y < ( -1/2 ) x + 3

dirty problem.

the diagram strongly suggests that we should only consider points located in the first quadrant.
if we only consider points in the first quadrant, then those points satisfy all three of inequalities (a), (d), (e).
to eliminate the former two, you have to consider points that lie beneath the given line but outside the first quadrant.

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the best way to do this problem is to write the equation of the line in slope-intercept form, as has been done by the poster above me.

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if you don't know how to do this, or you've forgotten how, then you can also plug in points.

the key is that you should PLUG IN POINTS FROM DIFFERENT AREAS OF THE GIVEN REGION.

try the point (6, -1), which is guaranteed to lie within the given region [since (6, 0) is on the line].
this point doesn't satisfy (b) or (c), so eliminate those.

try the point (-1, 3), which is guaranteed to lie within the given region [since (0, 3) is on the line].
this point doesn't satisfy (a) or (d), so eliminate those.

(e) is the last man standing.