Question 1003011
Use the elimination method to find all solutions of the system
x^2 + y^2 = 5
x^2 - y^2 = 1 
---------------Adding eliminates y^2, find x
2x^2 = 6
Divide both sides by 2
x^2 = 3
x = +{{{sqrt(3)}}}
and
x = -{{{sqrt(3)}}}
we know that x^2 = 3, substitute 3 for x^2 in the first equation
3 + y^2 = 5
y^2 = 5 - 3
y^2 = 2
y = +{{{sqrt(2)}}}
and
y = -{{{sqrt(2)}}}
:
The four solutions of the system are:
the one with x<0,y<0 is
x=-{{{sqrt(3)}}}
y=-{{{sqrt(2)}}}
the one with x<0,y>0 is
x=-{{{sqrt(3)}}}
y=+{{{sqrt(2)}}}
the one with x>0,y<0 is
x=+{{{sqrt(3)}}}
y=-{{{sqrt(2)}}}

the one with x>0,y>0 is
x = {{{sqrt(3)}}}
y = {{{sqrt(2)}}}