Answer:
Alright, here we go.
11. Those are vertical angles. All vertical angles are equal.
6x + 7 = 8x - 17
2x = 24
x = 12
12. Those are linear angles (adjacent angles on a line). Linear angles = 180.
(11x - 15) + (5x - 13) = 180
16x - 28 = 180
16x = 208
x = 13
13. If BD is perpendicular to AC, that means ∠ABD and ∠CBD = 90°. ∠CBD is made up of angle DBE, which equals (2x - 1)° and angle CBE, which equals (5x - 42)°.
(5x - 42) + (2x - 1) = 90
7x - 43 = 90
7x = 133
x = 19
14. Angles PQS and SQR are equal and halves of angle PQR because of bisector QS. If PQR = 82°, angle PQS = 41°
10x + 1 = 41
10x = 40
x = 4
15. If the bottom angle is (x + 10)°, then the top angle is also (x + 10)° because they're vertical angles. Angle (x + 10) and angle (10x - 61) are linear angles, so they equal 180.
(x + 10) + (10x - 61) = 180
11x - 51 = 180
11x = 231
x = 21
As for y, insert 21 as x for (10x - 61).
10(21) - 61
210 - 61
149
(18y + 5) should also equal 149° bc of vertical angles.
18y + 5 = 149
18y = 144
y = 8
x = 21
y = 8
(16 is crossed off so I'm going to assume you're not doing that one.)
17. If angle MNQ = (8x + 12)°, and it's bisected by NP, we can say that angles MNP and PNQ are equal halves of (4x + 6)°. Now, we were told that angle PNQ = 78°.
4x + 6 = 78
4x = 72
x = 18
This means that angle MNQ = 78 x 2, which is 156.
Now to find y, we see that angles MNQ and RNM are linear angles, so we can say:
(3y - 9) + 156 = 180
3y - 9 = 24
3y = 33
y = 11
x = 18
y = 11
(This took a while.)