Answer:
[tex]\large\boxed{F=1\dfrac{4}{9}}[/tex]
Step-by-step explanation:
[tex]F-\dfrac{8}{9}=\dfrac{5}{9}\qquad\text{add}\ \dfrac{8}{9}\ \text{to both sides}\\\\F-\dfrac{8}{9}+\dfrac{8}{9}=\dfrac{5}{9}+\dfrac{8}{9}\\\\F=\dfrac{5+8}{9}\\\\F=\dfrac{13}{9}\\\\F=1\dfrac{4}{9}[/tex]
