Question

1. find the values of x and y.

(18y + 5)°
(10x - 61)°
(x + 10)°

1. find the values of x and y.

(2y + 5)°
(5x - 17)°
(3x - 11)°

Explanation:

Step1: Use vertical - angle property for problem 15

Vertical angles are equal. So, $10x−61=x + 10$.
Solve for $x$:
\[

$$\begin{align*} 10x−x&=10 + 61\\ 9x&=71\\ x&=\frac{71}{9}\approx7.89 \end{align*}$$

\]
Also, since the sum of angles around a point is $360^{\circ}$ and the other pair of vertical angles are equal, and we assume the angles are part of a linear - pair or vertical - angle relationship. If we consider the fact that the two given non - vertical angles are supplementary (a common case in such angle problems), we have no information about $y$ from this single equation for $x$. But if we assume the two non - vertical angles are vertical angles, then $18y + 5=10x−61$. Substitute $x = \frac{71}{9}$ into it:
\[

$$\begin{align*} 18y+5&=10\times\frac{71}{9}-61\\ 18y+5&=\frac{710}{9}-\frac{549}{9}\\ 18y+5&=\frac{710 - 549}{9}\\ 18y+5&=\frac{161}{9}\\ 18y&=\frac{161}{9}-5\\ 18y&=\frac{161 - 45}{9}\\ 18y&=\frac{116}{9}\\ y&=\frac{116}{9\times18}=\frac{58}{81}\approx0.72 \end{align*}$$

\]

Step2: Use angle - sum property for problem 16

The sum of angles on a straight line is $180^{\circ}$. For the angles $(5x−17)^{\circ}$ and $(3x - 11)^{\circ}$, we have $(5x−17)+(3x - 11)=180$.
\[

$$\begin{align*} 5x+3x&=180 + 17+11\\ 8x&=208\\ x&=26 \end{align*}$$

\]
Since the angle with measure $(2y + 5)^{\circ}$ and one of the other angles (say the one adjacent to it on a straight - line or in a right - angle relationship) are related. If we assume the angle $(2y + 5)^{\circ}$ and the angle formed by the right - angle and one of the $x$ - related angles are complementary. Let's assume the right - angle is part of the angle - sum situation. Then, if we consider the fact that the sum of angles around the intersection point is relevant. If we assume the angle $(2y + 5)^{\circ}$ and the angle $(3x - 11)^{\circ}$ are complementary (since there is a right - angle in the figure), substitute $x = 26$ into the equation $2y+5+3x - 11 = 90$.
\[

$$\begin{align*} 2y+5+3\times26-11&=90\\ 2y+5 + 78-11&=90\\ 2y+72&=90\\ 2y&=90 - 72\\ 2y&=18\\ y&=9 \end{align*}$$

\]

Answer:

For problem 15: $x=\frac{71}{9},y=\frac{58}{81}$
For problem 16: $x = 26,y = 9$