please answer this question
Answers
Step-by-step explanation:
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.