If AB' is the image of AB after a reflection in AC, then:
[tex]\begin{gathered} m\angle BAC+m\angle CAB^{\prime}=m\angle BAB^{\prime} \\ \text{Where:} \\ m\angle BAC=m\angle CAB^{\prime}=4x+7 \\ m\angle BAB^{\prime}=64 \\ 4x+7+4x+7=64 \\ 8x+14=64 \\ 8x=64-14 \\ 8x=50 \\ x=\frac{50}{8} \\ x=\frac{24}{4}=6.25 \end{gathered}[/tex]
