SOLUTION: 27x^3+y^3=35; xy=2; find 3x+y

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: 27x^3+y^3=35; xy=2; find 3x+y      Log On


   

Question 998683: 27x^3+y^3=35; xy=2; find 3x+y
Answer by ikleyn(52784) About Me  (Show Source):
You can put this solution on YOUR website!
.
27x%5E3%2By%5E3 = 35;
xy = 2.

Find 3x+y.
---------------------------------------

Express  y  from the second equation  y = 2%2Fx  and  substitute into the first one.  You will get

27x%5E3+%2B+%288%2Fx%5E3%29 = 35,

Multiply both sides by  x%5E3.  You will get

27x%5E6+%2B+8+ = 35x%5E3,     or

27x%5E6+-+35x%5E3+%2B+8 = 0.

Introduce new variable  z = x%5E3.  You will get a quadratic equation

27z%5E2+-+35z+%2B+8 = 0.

Now,  apply the quadratic formula to solve it.  You will get

z = %2835+%2B-+sqrt%2835%5E2+-+4%2A35%2A8%29%29%2F54 = %2835+%2B-+19%29%2F54.

So,  z%5B1%5D = 1     ----->  x = 1,  y = 2.

z%5B2%5D = 8%2F27     ----->  x = 2%2F3,  y = 3.

Answer.  Two solutions: 1)  x = 1,  y = 2;   and  2)  x = 2%2F3,  y = 3.