91Ó°ÊÓ

Arrowheads denotes line segment $\stackrel{¯}{AB},\stackrel{¯}{CD}$, and $\stackrel{¯}{EF}$are parallel

Line segment $\stackrel{¯}{BD}$is transversal to the parallel lines $\stackrel{¯}{AB},\stackrel{¯}{CD}$

Line segment $\stackrel{¯}{DF}$is perpendicular to the parallel lines$\stackrel{¯}{CD},\stackrel{¯}{EF}$

$∠ABD$and $∠BDC$forms a pair of same-side interior angles

Theorem 3-3: If two parallel lines are cut by a transversal then the same-side interior angles are supplementary.

$m∠ABC+m∠BDC=180$

$70+4x−2y=180$

$4x−2y=180−70$

$4x−2y=110\left(1\right)$

$∠CDF$and $∠DFE$forms a pair of same-side interior angles

Theorem 3-3: If two parallel lines are cut by a transversal then the same-side interior angles are supplementary.

$m∠CDF+m∠DFE=180$

$4x+2y+50=180$

$4x+2y=180−50$

$4x+2y=130\left(2\right)$

$$\begin{gathered} 4x-2y=110 \\ \underline{4x-2y=130} \\ 8x=110 \\ x=30 \end{gathered}$$

$$\begin{gathered} 4\left(30\right)−2y=110 \\ −2y=110−120 \\ y=\frac{−10}{−2} \\ y=5 \end{gathered}$$

Thus, $x=30$and $y=5$