Question

1. given the diagram below, find y and z.

a. solve for y.
y = 0
b. find m∠aec
c. find m∠ceb
d. solve for z.

Explanation:

Step1: Use vertical - angle property

Vertical angles are equal. So, $3y + 25=9y - 11$.

Step2: Solve the equation for y

Subtract $3y$ from both sides: $25 = 6y-11$. Then add 11 to both sides: $36 = 6y$. Divide both sides by 6, we get $y = 6$.

Step3: Find $m\angle AEC$

$\angle AEC$ and $(9y - 11)^{\circ}$ are supplementary. First, substitute $y = 6$ into $9y - 11$, we have $9\times6-11=54 - 11 = 43^{\circ}$. Then $m\angle AEC=180-(9y - 11)=180 - 43=137^{\circ}$.

Step4: Find $m\angle CEB$

$\angle CEB$ and $(9y - 11)^{\circ}$ are vertical - angles. Substitute $y = 6$, $m\angle CEB=9y - 11=9\times6-11 = 43^{\circ}$.

Step5: Use linear - pair property to solve for z

$\angle AED$ and $\angle CEB$ are vertical - angles, so $\angle AED=\angle CEB = 43^{\circ}$. Also, $\angle AED=(12z - 7)^{\circ}$. Then set up the equation $12z-7 = 43$. Add 7 to both sides: $12z=50$. Divide both sides by 12: $z=\frac{25}{6}$.

Answer:

a. $y = 6$
b. $m\angle AEC = 137^{\circ}$
c. $m\angle CEB = 43^{\circ}$
d. $z=\frac{25}{6}$