Answer
$(-1/2, -3/2)$
$(1/2, 3/2)$
Work Step by Step
Given the system of equations:
1. $x^2 + xy = 1$
2. $xy + y^2 = 3$
The question asks to find all real solutions.
Add equations 1 and 2
$x^2 + xy = 1$
+ ($xy + y^2 = 3$)
$x^2 + 2xy + y^2 = 4$
$(x + y)^2 = 4$
$x + y = +/- 2$
$x + y = -2$
$x + y = 2$
Thus $x = -2 - y$ and $x = 2 - y$
Substitute the values into the equation
$(-2 - y) y + y^2 = 3$
$-2y = 3$
$y = -3/2$
$x = -2 - (-3/2)$
$x = -1/2$
$(2 - y) y + y^2 = 3$
$2y = 3$
$y = 3/2$
$x = 2 - 3/2$
$x = 1/2$
All Solutions:
$(-1/2, -3/2)$
$(1/2, 3/2)$
