Question

1. what is the value of x? 19 21 38 42

Explanation:

Step1: Identify vertical angles

$\angle AEB = \angle CED$ (vertical angles are equal)

Step2: Sum angles in triangle AEB

$\angle AEB = 180^\circ - 42^\circ - 2x^\circ = 138^\circ - 2x^\circ$

Step3: Sum angles in triangle CED

$\angle CED = 180^\circ - 38^\circ - (6y+14)^\circ = 128^\circ - 6y^\circ$

Step4: Set vertical angles equal

$138 - 2x = 128 - 6y \Rightarrow 2x - 6y = 10 \Rightarrow x - 3y = 5$
(Note: Assuming $\angle BAC = \angle BDC$ or $\angle ABD = \angle ACD$ is not needed—correction: Actually, $\angle ABE = \angle CDE$ (inscribed angles over arc AE) so $\angle ABE = 42^\circ = \angle CDE = 6y+14$)

Step5: Solve for y first

$6y + 14 = 42 \Rightarrow 6y = 28 \Rightarrow y = \frac{14}{3}$ (No, better: $\angle BEC = \angle AED$? Wait, $\angle BCA = \angle BDA$ so $38^\circ = 2x^\circ$? No, $\angle ABC = \angle ADC$ so $42^\circ = 6y+14 \Rightarrow 6y=28 \Rightarrow y=14/3$ no. Wait, $\angle BEC = 180 - 42 -38=100$, so $\angle AED=100$, then $\angle EAD + \angle EDA=80 \Rightarrow 2x +6y+14=80 \Rightarrow 2x+6y=66 \Rightarrow x+3y=33$. Earlier $x-3y=5$, add equations: $2x=38 \Rightarrow x=19$)

Step6: Solve for x

$x=19$

Answer:

19