SOLUTION: Solve the system of equations x^2 + y^2 - x - y = 18 xy + 2x + 2y = 26 I'm really having hard time at this part.

Algebra ->  Systems-of-equations -> SOLUTION: Solve the system of equations x^2 + y^2 - x - y = 18 xy + 2x + 2y = 26 I'm really having hard time at this part.      Log On


   

Question 983071: Solve the system of equations
x^2 + y^2 - x - y = 18
xy + 2x + 2y = 26
I'm really having hard time at this part.
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!

You need to solve the non-linear system of two equations in two unknowns

system%28x%5E2+%2B+y%5E2+-+x+-+y+=+18%2C%0D%0Axy+%2B+2x+%2B+2y+=+26%29.

Multiply the second equation by  2  (both sides)  and then add to the first one.  You will get

x%5E2+%2B+2xy+%2B+y%5E2 + 3%2A%28x%2By%29 = 18+%2B+2%2A26,   or

%28x+%2B+y%29%5E2 + 3%2A%28x%2By%29 - 70 = 0.

The last equation is quadratic for  (x + y)  and has the roots

a) x + y = -10   and   b) x + y = 7.

So,  combining this with the second equation of the original system, you should solve two systems

a) system%28x+%2B+y+=+-10%2C%0D%0Axy+%2B+2%2A%28-10%29+=+26%29     and     b) system%28x+%2B+y+=+7%2C%0D%0Axy+%2B+2%2A7+=+26%29.

or

a) system%28x+%2B+y+=+-10%2C%0D%0Axy+=+46%29     and     b) system%28x+%2B+y+=+7%2C%0D%0Axy+=+40%29.

The system  a)  has no real solutions.

The system  b)  has two solutions  x=4, y=3  and  x=3, y=4.

Answer.  Two solutions are  x=4, y=3  and  x=3, y=4.


Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
system%28x%5E2+%2B+y%5E2+-+x+-+y+=+18%2C%0D%0Axy+%2B+2x+%2B+2y+=+26%29

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%5E2+%2B+y%5E2+-+3x+-+3y+=+8%2C++%0D%0Axy+%2B+2x+%2B+2y+=+24%29

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+%2B+2x+%2B+2y+=+24
xy%2B2y=24-2x
y%28x%2B2%29=24-2x
y+=+%2824-2x%29%2F%28x%2B2%29 

Substitute in the first equation:

x%5E2+%2B+%28%2824-2x%29%2F%28x%2B2%29%29%5E2+-+3x+-+3%28%2824-2x%29%2F%28x%2B2%29%29+=+8

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



Get binomials in descending order:







x%5E4%2Bx%5E3-6x%5E2-200x%2B400+=+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

%28x-2%29%28x-5%29%28x%5E2%2B8x%2B40%29+=+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+=+%28-8+%2B-+sqrt%2864-160+%29%29%2F2+

x+=+%28-8+%2B-+sqrt%28-96+%29%29%2F2+

x+=+%28-8+%2B-+i%2Asqrt%2896+%29%29%2F2+

x+=+%28-8+%2B-+i%2Asqrt%2816%2A6+%29%29%2F2+

x+=+%28-8+%2B-+4i%2Asqrt%286%29%29%2F2+

x+=+%282%28-4+%2B-+i%2Asqrt%286%29%29%29%2F2+

x+=+-4+%2B-+i%2Asqrt%286%29+

So the other pair of solutions, by symmetry, are

if x+=+-4+%2B+i%2Asqrt%286%29+, then y+=+-4+-+i%2Asqrt%286%29+

and 

if x+=+-4+-+i%2Asqrt%286%29+, then y+=+-4+%2B+i%2Asqrt%286%29+

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

Edwin