Answer
The answer is $(x,y)=\{(-3,4)$ and $(4,-3)\}$.
Work Step by Step
The given equations are
$x^2+y^2=25$ ... (1)
$x+y=1$ ... (2)
Square the equation (2).
$(x+y)^2=1^2$
Clear the parentheses.
$x^2+y^2+2xy=1$
Substitute values from equation (1).
$25+2xy=1$
$2xy=1-25$
$2xy=-24$
$xy=-\frac{24}{2}$
$x=-\frac{12}{y}$ Substitute into equation (2).
$-\frac{12}{y}+y=1$
Multiply the equation by $y$
$-12+y^2=y$
$y^2-y-12=0$
Factor the equation.
$y^2-4y+3y-12=0$
$y(y-4)+3(y-4)=0$
$(y-4)(y+3)=0$
The possible values of $y$ is
$y=4,-3$ Substitute into equation (2).
$x+4=1$
$x=1-4$
$x=-3$
and $x-3=1$
$x=1+3$
$x=4$
The solution set is
$\{(-3,4)$ and $(4,-3)\}$.
