[tex]F=\dfrac{9}{5}C+32\\\\A.\\\dfrac{9}{5}C+32=F\ \ \ |-32\\\\\dfrac{9}{5}C=F-32\ \ \ |\cdot\dfrac{5}{9}\\\\C=\dfrac{5}{9}(F-32)[/tex]
[tex]B.\\F=212\to C=\dfrac{5}{9}(212-32)=\dfrac{5}{9}\cdot180=100\\\\212^oF=100^oC[/tex]
[tex]C.\\F=80\to C=\dfrac{5}{9}(80-32)=\dfrac{5}{9}\cdot48\approx26.7\\\\80^oF\approx26.7^oC[/tex]