SOLUTION: (x+y)(x^2+y^2)=40 (x-y)(x^2-y^2)=16 x=? y=?

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Question 1105564: (x+y)(x^2+y^2)=40
(x-y)(x^2-y^2)=16
x=?
y=?
Found 3 solutions by Alan3354, ikleyn, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(x+y)(x^2+y^2)=40
(x-y)(x^2-y^2)=16
x=?
y=?
---------------
(3,1) and (1,3)

Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.
(x+y)(x^2+y^2)=40    (1)
(x-y)(x^2-y^2)=16    (2)

=================>

x^3 + xy^2 + yx^2 + y^3 = 40   (1')
x^3 - xy^2 - yx^2 + y^3 = 16   (2')


Add (1') and {2') (both sides). You will get

2(x^3 + y^3) = 56,     or  x^3 + y^3 = 28.     (3)


Subtract (2') from (1'). You will get

2(xy^2 + yx^2) = 24,   or  xy^2 + x^2y = 12.   (4)


Then

x^3 + 3x^2y + 3xy^2 + y^3 = 28 + 3*12 = 64,   or

(x+y)^3 = 64,   which implies   

x + y = 4       (5)


Next, from (4) you have

xy*(x+y) = 12,  and,  replacing here x+y by 4 (due to (5)), you get

xy*4 = 12,   or

xy = 12%2F4 = 3.    (6)


From (5) and (6) you can easily get the 


Answer.  (x,y) = (1,3)   or  (x,y) = (3,1).

Solved.




Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
He's no more help than the back of the book. LOL
She didn't give the complex solutions.

system%28%28x%2By%29%28x%5E2%2By%5E2%29=40%2C%0D%0A%28x-y%29%28x%5E2-y%5E2%29=16%29

Notice that if we interchange x and y, the system of equations
simplifies to the same system, so if (x,y)=(p,q) is a solution,
then so is (x,y)=(q,p) 

Let y = kx



system%28x%281%2Bk%29x%5E2%281%2Bk%5E2%29=40%2C%0D%0Ax%281-k%29x%5E2%281-k%5E2%29=16%29

system%28x%5E3%281%2Bk%29%281%2Bk%5E2%29=40%2C%0D%0Ax%5E3%281-k%29%281-k%5E2%29=16%29

Solve each for x³



Equate the expressions for x³

40%2F%281%2Bk%29%281%2Bk%5E2%29=16%2F%281-k%29%281-k%5E2%29

Divide both sides by 8

5%2F%281%2Bk%29%281%2Bk%5E2%29=2%2F%281-k%29%281-k%5E2%29

Cross-multiply:

2%281%2Bk%29%281%2Bk%5E2%29=5%281-k%29%281-k%5E2%29

Factor 1-k² as the difference of squares:

2%281%2Bk%29%281%2Bk%5E2%29=5%281-k%29%281-k%29%281%2Bk%29

Get 0 on the right

2%281%2Bk%29%281%2Bk%5E2%29-5%281-k%29%281-k%29%281%2Bk%29=0

Factor out common factor (1+k)

%281%2Bk%29%282%281%2Bk%5E2%29%5E%22%22-5%281-k%29%281-k%29%5E%22%22%29=0

%281%2Bk%29%282%2B2k%5E2-5%281-2k%2Bk%5E2%29%29=0

%281%2Bk%29%282%2B2k%5E2-5%2B10k-5k%5E2%29=0

%281%2Bk%29%28-3k%5E2+%2B+10k+-+3%29=0

%281%2Bk%29%28-3k%5E2+%2B+10k+-+3%29=0

Use the zero factor property:

1+k = 0;    -3k²+10k-3 = 0
  k = -1;    3k²-10k+3 = 0
           (3k-1)(k-3) = 0  
         3k-1 = 0;  k-3 = 0
           3k = 1;    k = 3
            k = 1/3

We substitute those three values for k in:

x%5E3%281-k%29%281-k%5E2%29=16

Substituting k=-1

x%5E3%281-%28-1%29%5E%22%22%29%281-%28-1%29%5E2%29=16
x%5E3%281%2B1%29%281-1%29=16
0=16, not possible:

Substituting k=1/3

x%5E3%281-%281%2F3%29%5E%22%22%29%281-%281%2F3%29%5E2%29=16
x%5E3%282%2F3%29%281-1%2F9%29=16 
x%5E3%282%2F3%29%288%2F9%29=16
x%5E3%2816%2F27%29=16
x%5E3%2816%29=16%2A27
x%5E3=27
x%5E3-27=0
%28x-3%29%28x%5E2%2B3x%2B9%29=0+}}}

Use zero-factor property:

x=3; x²+3x+9=0
     x+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A9%2A1+%29%29%2F%282%2A1%29+
     x+=+%28-3+%2B-+sqrt%289-36%29%29%2F2+
     x+=+%28-3+%2B-+sqrt%28-27%29%29%2F2+
     x+=+%28-3+%2B-+i%2Asqrt%2827%29%29%2F2+
     x+=+%28-3+%2B-+i%2Asqrt%289%2A3%29%29%2F2+
     x+=+%28-3+%2B-+3i%2Asqrt%283%29%29%2F2+

And since y = kx = (1/3)x

when x = 3, y = (1/3)(1) = 1

So one solution is (x,y) = (3,1), and by swapping x and y, we
have

 (x,y) = (1,3)

and when x+=+%28-3+%2B-+3i%2Asqrt%283%29%29%2F2+,  y+=+%281%2F3%29%28%28-3+%2B-+3i%2Asqrt%283%29%29%2F2%29+=+%28-1+%2B-+i%2Asqrt%283%29%29%2F2+

So two more solutions are



and



And two more solutions by swapping x and y are



and



Edwin