Solve-the-system-of-equations-x-2-y-2-2xy-x-y-1-x-y-x-2-y-in-real-numbers-x-y-

Question Number 122002 by liberty last updated on 13/Nov/20

Solve the system of equations

{\begin{cases} x^{2} + y^{2} + \frac{2 xy}{x + y} = 1 \\ \sqrt{x + y} = x^{2} - y \end{cases} in real numbers x, y .

Answered by MJS_new last updated on 13/Nov/20

\sqrt{x + y} = x^{2} - y

squaring & transforming

y^{2} - (2 x^{2} + 1) y + (x^{4} - x) = 0

\Rightarrow (a) y = x^{2} - x \lor (b) y = x^{2} + x + 1

[we have some limitations but {let}^{'} s just go on

and see what will happen]

inserting in (1) we get

(a) x^{6} - 2 x^{5} + 2 x^{4} + 2 x^{3} - 3 x^{2} = 0

(b) x^{6} + 4 x^{5} + 9 x^{4} + 14 x^{3} + 10 x^{2} + 4 x = 0

\Rightarrow

(a) x^{2} (x - 1) (x + 1) (x^{2} - 2 x + 3) = 0

(b) x (x + 2) (x^{4} + 2 x^{3} + 5 x^{2} + 4 x + 2) = 0

\Rightarrow

(a) x = - 1 \lor x = 0 \lor x = 1

\Rightarrow y = 2^{}^{} \lor y = 0 \lor y = 0

(b) x = - 2 \lor x = 0

\Rightarrow y = 3 \lor y = 1

testing all pairs we get

(x ∣ y) = (1 ∣ 0) \lor (- 2 ∣ 3)

Commented by liberty last updated on 13/Nov/20

thank you sir

Answered by liberty last updated on 13/Nov/20

my way

multiply the first eq by x + y give

(x^{2} + y^{2}) (x + y) + 2 xy = x + y

adding x^{2} + y^{2} to both sides of this eq, we are

driven by standard manipulation to

nice factorization :

(x^{2} + y^{2}) (x + y) + {(x + y)}^{2} = (x^{2} + y^{2}) + (x + y)

(x^{2} + y^{2}) (x + y - 1) + (x + y) (x + y - 1) = 0

(x^{2} + y^{2} + x + y) (x + y - 1) = 0

The first factor cannot be zero because x + y > 0

so x + y - 1 must be zero . Inserting

y = 1 - x into second eq of the system we

find the two solutions (x, y) : (1, 0) and (- 2, 3) . ▴

Facebook Tweet Pin