평행선과 각의 성질을 이용한 미지수 값 구하기

문제

그림과 같이 두 직선 $l, m$이 평행하고, 두 점 $A, C$는 직선 $l$ 위에, 두 점 $B, D$는 직선 $m$ 위에 있습니다. 선분 $AD$와 $BC$의 교점을 $E$라고 할 때, $\angle DAC = (3x-15)^\circ$, $\angle ADB = (x+35)^\circ$, $\angle CBE = 50^\circ$, $\angle BCE = (y+10)^\circ$ 입니다. 이때, $y$의 값은?

\begin{tikzpicture}
\draw[thick] (-3,2) -- (3,2) node[right] {$l$};
\draw[thick] (-3,-2) -- (3,-2) node[right] {$m$};

\coordinate (A) at (-1.5,2);
\coordinate (C) at (1.5,2);
\coordinate (B) at (-2,-2);
\coordinate (D) at (2,-2);

\
ode[above left] at (A) {$A$};
\
ode[above right] at (C) {$C$};
\
ode[below left] at (B) {$B$};
\
ode[below right] at (D) {$D$};

\draw[thick] (A) -- (D);
\draw[thick] (C) -- (B);

\coordinate (E) at (intersection of A--D and C--B);
\
ode[above right] at (E) {$E$};

% Angles
% Angle DAC
\draw[dashed] (A) ++(1.5,0) -- (A) ++(0.5,0);
\draw[thick] (A) -- (D);
\pic[draw,angle radius=10mm, "{$ (3x-15)^\circ $}", angle eccentricity=1.2] {angle = D--A--C};

% Angle ADB
\draw[dashed] (D) ++(-1.5,0) -- (D) ++(-0.5,0);
\draw[thick] (A) -- (D);
\pic[draw,angle radius=10mm, "{$ (x+35)^\circ $}", angle eccentricity=1.2] {angle = A--D--B};

% Angle CBE
\draw[dashed] (B) ++(1.5,0) -- (B) ++(0.5,0);
\draw[thick] (C) -- (B);
\pic[draw,angle radius=10mm, "{$ 50^\circ $}", angle eccentricity=1.2] {angle = E--B--D};

% Angle BCE
\draw[dashed] (C) ++(-1.5,0) -- (C) ++(-0.5,0);
\draw[thick] (C) -- (B);
\pic[draw,angle radius=10mm, "{$ (y+10)^\circ $}", angle eccentricity=1.2] {angle = E--C--A};

\end{tikzpicture}