Answer :
Given:
[tex](11x-11)^{\circ}and(16x-7)^{\circ}\text{ forms an complementary angles.}[/tex][tex]\begin{gathered} 11x-11+16x-7=90 \\ 27x-18=90 \\ 27x=90+18 \\ 27x=108 \\ x=\frac{108}{27} \\ x=4 \end{gathered}[/tex]Given:
[tex](11x-11)^{\circ}and(16x-7)^{\circ}\text{ forms an complementary angles.}[/tex][tex]\begin{gathered} 11x-11+16x-7=90 \\ 27x-18=90 \\ 27x=90+18 \\ 27x=108 \\ x=\frac{108}{27} \\ x=4 \end{gathered}[/tex]