Please help don’t remember what to do

  (4)  (4, 0)

Step-by-step explanation:

To find out if a given point is a solution, try it and see if the inequalities are true with those values.

Personally, I don't like to mess with fractions, so I would rewrite the first inequality as ...

  2y < x +8 . . . . . multiply the first inequality by 2

Now, we can try the given points to see if they work.

  (1) 2·3 < -5 +8   ⇒ 6 < 3 . . . . False

  (2) 2·4 < 0 +8   ⇒ 8 < 8 . . . . False

  (3) 2·(-5) < 3 +8   ⇒ -10 < 11 . . . . True, so we need to look at the second inequality

       -5 ≥ -(3) +1   ⇒ -5 ≥ -2 . . . . False

So far, we have ruled out the first three answer choices. We expect the last answer choice will satisfy both inequalities.

  (4) 2·0 < 4 +8   ⇒ 0 < 12 . . . . True

       0 ≥ -4 +1   ⇒ 0 ≥ -3 . . . . True

The last answer choice (4, 0) satisfies both inequalities, so is in their solution set.

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This can be confirmed by a graph. Points that satisfy both inequalities will be in the doubly-shaded area or on its solid boundary line. A point on a dashed boundary line is not part of the solution.