Question 983071
<pre>
{{{system(x^2 + y^2 - x - y = 18,
xy + 2x + 2y = 26)}}}

Instead of doing your problem for you, I will do one exactly 
in every detail step-by-step like your problem.  All you have
to do is use the problem below as a model. The problem I will
do is:

{{{x^2 + y^2 - 3x - 3y = 8,  
xy + 2x + 2y = 24)}}}

One thing we observe about this problem is that it is symmetrical
in x and y.  That is we have the same two equations if we interchange
x and y.  Therefore any time we get a pair of solutions for x, one will
be a solution for x and the other will be a solution for y.

Solve the second for y

{{{xy + 2x + 2y = 24}}}
{{{xy+2y=24-2x}}}
{{{y(x+2)=24-2x}}}
{{{y = (24-2x)/(x+2)}}} 

Substitute in the first equation:

{{{x^2 + ((24-2x)/(x+2))^2 - 3x - 3((24-2x)/(x+2)) = 8}}}

Multiply through by (x+2)², [but we know x can't be -2]

{{{x^2(x+2)^2 + (24-2x)^2 - 3x(x+2)^2 - 3(24-2x)(x+2) = 8(x+2)^2}}}

Get binomials in descending order:

{{{x^2(x+2)^2 + (-2x+24)^2 - 3x(x+2)^2 - 3(-2x+24)(x+2) = 8(x+2)^2}}}

{{{x^2(x^2+4x+4) + (4x^2-96x+576)-3x(x^2+4x+4)- 3(-2x^2+20x+48) = 8(x^2+4x+4)}}}

{{{x^4+4x^3+4x^2 + 4x^2-96x+576 - 3x^3-12x^2-12x +6x^2-60x-144 = 8x^2+32x+32}}}

{{{x^4+x^3-6x^2-200x+400 = 0}}}

Looking at the graph on a graphing calculator it appears that 
it has solutions 2 and 5 but it probably has a pair of conjugate
complex solutions too.  We use synthetic division with x=2

2 | 1 1 -6 -200  400
  |   2  6    0 -400 
    1 3  0 -200    0

5 | 1  3   0 -200
  |    5  40  200    
    1  8  40    0

So we have factored the polynomial as

{{{(x-2)(x-5)(x^2+8x+40) = 0}}}

So we have solutions x=2, x=5, 

By the symmetry, we know that when x=2, y=5
and when x=5, y=2.

we get the other solutions from
using the quadratic formula, since it won't factor.

{{{x^2+8x+40 = 0}}]

{{{x = (-8 +- sqrt(8^2-4*1*40 ))/(2*1) }}}

{{{x = (-8 +- sqrt(64-160 ))/2 }}}

{{{x = (-8 +- sqrt(-96 ))/2 }}}

{{{x = (-8 +- i*sqrt(96 ))/2 }}}

{{{x = (-8 +- i*sqrt(16*6 ))/2 }}}

{{{x = (-8 +- 4i*sqrt(6))/2 }}}

{{{x = (2(-4 +- i*sqrt(6)))/2 }}}

{{{x = -4 +- i*sqrt(6) }}}

So the other pair of solutions, by symmetry, are

if {{{x = -4 + i*sqrt(6) }}}, then {{{y = -4 - i*sqrt(6) }}}

and 

if {{{x = -4 - i*sqrt(6) }}}, then {{{y = -4 + i*sqrt(6) }}}

Now you do your problem exactly step-by step like I did this one.

Edwin</pre>