15.

3x + 7y = 4y + 14
3x + 3y = 14
x + y = 14/3

16.

4x – 2y + 22 + 18 ³ 50
4x – 2y ³ 10
2x – y ³ 5
Now add the inequality y > 7 to get
2x > 12.

\ x > 6

17.

Successful points are all points on segment FG, plus the portion of segment GH from –2 to 3. Thus the successful points are from –6 to +3, a distance of 9. Since total distance = 15, probability is 9/15 = 3/5.

18.

Equation (1): (x + y) + (y + 15) = 180, which simplifies to x + 2y = 165.
Equation (2): (y + 15) + (39 – x) = 90, which simplifies to –x + y = 36.
Add (1) to (2) to get 3y = 201, from which we conclude y = 67.
Back-substitute into either equation to get x = 31.

Now mÐTAN = 39 – x = 39 – 31 = 8, so mÐMAT = 180 – 8 = 172.

19.

1. ÐHOB @ ÐHAT (Given)
2. ÐHOA supp. ÐHOB, ÐHAO supp. ÐHAT (Diagram; def. str. Ð)
3. ÐHOA @ ÐHAO (Supps. of @ Ðs are @)
4. ray OG bisects ÐHOA, ray AG bisects ÐHAO (Given)
5. ÐHOG @ ÐHAG (Div. Prop.)
6. ÐBOG @ ÐTAG (Add. Prop. from steps 1 & 5)

20.

It is not always possible to take the complement of an angle. (Only acute angles have complements.) Thus, whenever the supplement of an angle is right or obtuse, that supplement is something for which a complement cannot be computed.

However, every angle has a supplement. Thus, the supplement of the complement is always possible (provided the complement itself exists, that is).