Use the elimination method to find all solutions of the system { x 2 + y 2 = 5 x 2 − y 2 = 4

Question

The four solutions of the system are:

the one with x < 0 , y < 0 is

x =

y =

the one with x < 0 , y > 0 is

x =

y =

the one with x > 0 , y < 0 is

x =

y =

the one with x > 0 , y > 0 is

x =

y =

Answer 1

x² + y² = 5

x² - y² = 4

2x² = 9

x² = 9 / 2

x = ±3√2 / 2

9 / 2 + y² = 5

y² = 5 - 9/2

y² = 1 / 2

y = ±√2 / 2

Answer 2

Let x^2 = v and y^2 = w

Then,

v + w = 5

5v - w = 4

v = 5 - w

5(5-w) - w = 4

25 - 5w - w = 4

21 = 6w

w = 7/2

v = 5 - w

v = 3/2

Then, x = ±√v = ±√(3/2) = ±√6/2

y = ±√w = ±√(7/2) = ±√14/2

x < 0, y < 0

x = -√6/2

y = -√14/2

x < 0, y > 0

x = -√6/2

y = √14/2

x > 0, y < 0

x = √6/2

y = -√14/2

x > 0, y > 0

x = √6/2

y = √14/2