[tex] \left \{ {{x-1/y=2/3} \atop {x^2+1/y^2=10/9}} ight. [/tex]

{(x-1) /y =  2/3{(x²+1) / y² = 10/9y≠0  , у²≠0{3(x-1) = 2y{ 9(x²+1) =10y²{ 3x-3  -2y=0{9x²+9  - 10y² =0{y = (3x- 3)/2      ⇒ y= 3/2 * (x-1) = 1.5(x-1){9x²-10y² +9=09x²-10*(1.5(x-1))² +9 =09х² - 10* (1,5х - 1,5)² +9=09х² - 10 *(2,25х²  - 2*1,5*1,5х + 2,25) +9=09х² - 22,5х²  + 45х  -22,5+9=0-13,5х²  +45х  - 13,5  = 0             |÷ (-45)0.3x² - x  + 0.3 =0D = 1  -  4*0.3*0.3= 1- 0.36=0.64= 0.8²x₁= (1+0.8)/(2*0.3)= 1.8/0.6=3x₂ = (1-0.8)/ 0.6 = 0.2/0.6 = 1/3y₁=1.5 (3-1) = 1.5*2 = 3y₂= 1.5 (1/3 - 1)  = 3/2  *  (-2/3) = -1Ответ: (3;3)  ;  (1/3 ;-1)