If $x+y=4$ and $1/x+1/y=4/3$, then $x^2+y^2$ equals:

Combine the second condition: $1/x+1/y=(x+y)/(xy)=4/(xy)=4/3$, so $xy=3$. Then $x^2+y^2=(x+y{)}^2-2xy=16-6=10$. (16 forgets the $-2xy$ term; 13 subtracts $xy$ once instead of twice; 22 adds $2xy$.) The pair is $x,y=1$, 3: indeed 1 + 1/3 = 4/3 and 1 + 9 = 10.