A parallelogram has two diagonals that intersect at their midpoint, therefore:
[tex]\begin{gathered} JO=LO \\ so\colon \\ 5x+11=6x-3 \\ solve_{\text{ }}for_{\text{ }}x\colon \\ 11+3=6x-5x \\ 14=x \\ x=14 \end{gathered}[/tex]
in a parallelogram the angles of any two contiguous vertices are supplementary, therefore:
[tex]\begin{gathered} m\angle B+m\angle C=180 \\ 2x+15+x=180 \\ solve_{\text{ }}for_{\text{ }}x \\ 3x=180-15 \\ 3x=165 \\ x=\frac{165}{3} \\ x=55 \end{gathered}[/tex]
