Question 1048631
.
Solve the system and graph the curves:

x^2+y^2 = 2 ;and
x-y = 4
~~~~~~~~~~~~~~~~~~~~


<pre>
{{{x^2+y^2}}} = 2,     (1)
x-y = 4          (2)

Express x = 4 + y from (2) and substitute it into (1). You will get

{{{(4 +y)^2 + y^2}}} = 2,  or

{{{16 + 8y + y^2 + y^2}}} = 2,  or

{{{2y^2 + 8y + 14}}} = 0,  or

{{{y^2 + 4y + 7}}} = 0.

Apply the quadratic formula. You will get

{{{y[1,2]}}} = {{{(-4 +- sqrt((-4)^2 - 4*7))/2}}} = {{{(-4 +- sqrt(-12))/2}}}.

There are no real solutions.
</pre>

See the plot

{{{graph( 330, 330, -4.5, 4.5, -4.5, 4.5,
          sqrt(2-x^2), -sqrt(2-x^2), x-4
)}}}


Plots y = {{{sqrt(2-x^2)}}}, y = {{{-sqrt(2-x^2)}}} (red+green) and y = x-4 (blue).