Solve the equations x 2 y + y 2 x = 9 2 , 3 x + y = 1 .

Given x 2 y + y 2 x = 9 2 ⇒ x 3 + y 3 = 9 2 x y . . . i And, 3 x + y = 1 ⇒ x + y = 3 . . . ii Now, from i , we have x 3 + y 3 = 9 2 x y ⇒ ( x + y ) 3 - 3 x y ( x + y ) = 9 2 x y ⇒ 27 - 9 x y = 9 2 x y ; By ii ⇒ 27 2 x y = 27 ⇒ x y = 2 . . . iii Using ii and iii , we have x 3 - x = 2 ⇒ 3 x - x 2 = 2 ⇒ x 2 - 3 x + 2 = 0 ⇒ x 2 - 2 x - x + 2 = 0 ⇒ x x - 2 - 1 x - 2 = 0 ⇒ x - 1 x - 2 = 0 ⇒ x = 2 , 1 Substituting these values in ii , we get y = 1 , 2

Solve the equations x 2 y + y 2 x = 9 2 , 3 x + y = 1 .

Given x 2 y + y 2 x = 9 2 ⇒ x 3 + y 3 = 9 2 x y . . . i And, 3 x + y = 1 ⇒ x + y = 3 . . . ii Now, from i , we have x 3 + y 3 = 9 2 x y ⇒ ( x + y ) 3 - 3 x y ( x + y ) = 9 2 x y ⇒ 27 - 9 x y = 9 2 x y ; By ii ⇒ 27 2 x y = 27 ⇒ x y = 2 . . . iii Using ii and iii , we have x 3 - x = 2 ⇒ 3 x - x 2 = 2 ⇒ x 2 - 3 x + 2 = 0 ⇒ x 2 - 2 x - x + 2 = 0 ⇒ x x - 2 - 1 x - 2 = 0 ⇒ x - 1 x - 2 = 0 ⇒ x = 2 , 1 Substituting these values in ii , we get y = 1 , 2

Solve the equations x 2 + y 5 = 5 , 2 x + 5 y = 5 6 .

Given x 2 + y 5 = 5 ⇒ 5 x + 2 y = 50 . . . i And, 2 x + 5 y = 5 6 ⇒ 5 x + 2 y = 5 6 x y . . . ii By i and ii , we have 5 6 x y = 50 ⇒ x y = 60 ⇒ y = 60 x . . . iii Putting this value of y in ( i ) , we get 5 x + 2 60 x = 50 ⇒ x + 24 x = 10 ⇒ x 2 - 10 x + 24 = 0 ⇒ x 2 - 4 x - 6 x + 24 = 0 ⇒ x x - 4 - 6 x - 4 = 0 ⇒ ( x - 6 ) ( x - 4 ) = 0 ⇒ x = 6 , 4 Substituting these values in iii , we get y = 10 , 15

Solve the equations x + y + x - y = 4 , x 2 - y 2 = 9 .

Given equations are x + y + x - y = 4 . . . i x 2 - y 2 = 9 . . . i i From i i , we have ( x - y ) ( x + y ) = 9 ⇒ ( x - y ) ( x + y ) = 3 Again we have x + y - x - y 2 = x + y + x - y 2 - 4 ( x + y ) ( x - y ) = 16 - 12 = 4 x + y - x - y = ± 2 . . . i i i Now from i and i i i on addition, we have 2 x + y = 6 or 2 x + y = 2 ⇒ x + y = 9 or x + y = 1 . . . i v And on subtraction, we have 2 x - y = 2 or 2 x - y = 6 ⇒ x - y = 1 or x - y = 9 . . . v Solving i v & v , we get x = 5 Hence, y = ± 4 .

Solve the equations y + x 2 - 1 = 2 , x + 1 - x - 1 = y .

The two equations are y + x 2 - 1 = 2 . . . i x + 1 - x - 1 = y . . . i i Squaring i i , we have x + 1 + x - 1 - 2 x 2 - 1 = y ⇒ 2 x - 2 x 2 - 1 = y . . . i i i Putting this value of y in i , we have 2 x - 2 x 2 - 1 + x 2 - 1 = 2 ⇒ 2 x - x 2 - 1 = 2 ⇒ x 2 - 1 = 2 x - 2 Squaring both sides, we have x 2 - 1 = 4 x 2 - 8 x + 4 ⇒ 3 x 2 - 8 x + 5 = 0 ⇒ 3 x 2 - 3 x - 5 x + 5 = 0 ⇒ 3 x x - 1 - 5 x - 1 = 0 ⇒ ( 3 x - 5 ) ( x - 1 ) = 0 ⇒ x = 5 3 , 1 Substituting these values in i i i , we get y = 2 3 , 2

Solve the equations x y + y x = 10 3 , x + y = 10 .

The given equations are x y + y x = 10 3 . . . i x + y = 10 . . . i i From i , we have x + y x y = 10 3 ⇒ 10 x y = 10 3 ⇒ x y = 3 ⇒ x y = 9 Now, ( x - y ) 2 = ( x + y ) 2 - 4 x y = 100 - 36 = 64 ∴ x - y = 8 . . . i i i x - y = - 8 . . . i v Now by i and i i i , we have 2 x = 18 ⇒ x = 9 ∴ y = 1 And, by i and i v , we have 2 x = 2 ⇒ x = 1 ⇒ y = 9

Solve the equations x - y x + y + x + y x - y = 17 4 , x 2 + y 2 = 706 .

Given x - y x + y + x + y x - y = 17 4 . . . i x 2 + y 2 = 706 . . . i i From i , we have ( x - y ) 2 + ( x + y ) 2 ( x - y ) = 17 4 ⇒ 2 ( x + y ) ( x - y ) = 17 4 ⇒ 8 x + 8 y = 17 x - 17 y ⇒ 9 x = 25 y ⇒ y = 9 25 x . . . i i i Hence from i i , we have x 2 + 9 25 x 2 = 706 ⇒ 706 625 x 2 = 706 ⇒ x 2 = 625 ⇒ x = ± 25 ∴ y = ± 9