findx and y5x(1+1/x2 +y2)=12 and 5y( 1-1/x2+y2)=4

5x(1 + ¹/x² + y²) = 12------(I),

5y(1 — ¹/x² + y²) = 4--------(II).

MAKING THE COEFFICIENT, 5 THE SUBJECT OF THE FORMULA:

From eqn I,

5 = ¹²/x(x² +y²/x² +y²+1) -----III.

From eqnII,

5 = ⁴/y(x² + y²/x² + y²—1)----IV.

SUBST. INTEGER (5) FROM Eqn III INTO Eqn IV.

I.e

¹²/x(x²+y²/x²+y²+1) = ⁴/y(x²+y²/x²+y² —1) ,

OR

12/x ÷ 4/y = x²+y²/x²+y²-1 ÷ x²+y²/x²+y² + 1

OR

12/x ( y/4) = x² + y²/x²+y²—1 (x²+y²+1/x² + y²).

This reduces to:

3y/x = x² + y² + 1/x²+y²—1,

as x² + y² cancels each other.

Now, by comparison,

3y = x² + y² + 1------------(V),

x = x² + y² — 1-----------(VI).

SUBTRACT Eqn V & VI:

3y — x = 2 , or

x = 3y — 2 -------VII.

ALSO, ADD Eqn V & VI:

3y + x = 2x² + 2y² ---------VIII.

SUBST. x from Eqn VII into

Eqn VIII:

3y + 3y — 2 = 2(3y — 2)² + 2y²,

6y — 2 = 2(9y² —12y + 4) + 2y²,

6y — 2 = 18y² — 24y + 8 + 2y²,

6y — 2 = 20y² — 24y + 8.

Thus, 20y² — 30y + 10 = 0,

Or 2y² — 3y + 1 = 0.

Factorizing:

(2y — 1)(y — 1) = 0,

and y = ½, 0r 1.

TESTING FOR REAL

SOLUTIONS:

Subst. y = ½ in EqnII above, gives :

32x² — 20x² + 23 = 0;

12x² = — 23;

x² = —23/12;

x = √(—23/12) , Complex number

Also, Subst. y = 1 in EqnII above,

5(1)[1 — ¹/x² + 1] = 4;

5[x² + 1 — 1/x² + 1] = 4;

5(x²/x² + 1) = 4;

5x² = 4x² + 4;

5x² — 4x² — 4 = 0;

x² — 4 = 0;

x² = 4, or x = √4 = 2.

Hence, solutions are

x = 2, y = 1.

Regards

Arun (askIITians forum expert)