Question 765924
1.


Let's call the two numbers x and y. Then 
{{{x^2+y^2=4}}} and 
{{{x^2-y^2=1}}}


Now we can see that, by subtracting the second equation from the first one, that
{{{2y^2=3}}}
So
{{{y^2=3/2}}}
And
{{{y=sqrt(3/2)}}} or {{{y=-sqrt(3/2)}}}

Substituting this into the first equation gives us
{{{x^2+3/2=4}}}
So
{{{x^2=5/2}}}
And therefore
{{{x=sqrt(5/2)}}} or {{{x=-sqrt(5/2)}}}

2.


Same idea here. We're calling the two numbers x and y. We get
{{{x^2+2y^2=6}}}
{{{2x-y=2}}}
Now the problem is a bit trickier though. We can't really find the value of either x or y immediately, so we're going to use substitution.


According to the second equation, 
{{{2x-y=2}}}
We can rewrite this as
{{{y=2x-2}}}
Now we substitute this 'implicit' value of y into our first equation:

{{{x^2+2(2x-2)^2=6}}}
We'll make that a bit nicer:
{{{x^2+2(2x-2)^2=x^2+2(4x^2-8x+4)=x^2+8x^2-16x+8=9x^2-16x+8=6}}}
So,
{{{9x^2-16x+2=0}}}
Now, that's an equation we can solve. With the trusty old
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
We get 
{{{x=(16+-sqrt(256-4*9*2))/(2*9)=(16+-sqrt(256-72))/18=(16+-sqrt(184))/18}}}
Dividing the numerator and denominator by two gives us
{{{x=(1/9)*(8+-sqrt(46))}}}
And so
{{{y=2x-2=2*(1/9)*(8+-sqrt(46))-2}}}