Question

(i) The number x is 2 more than the number y. If the sum of the squares of x and y is 34; find the product of x and y.
(ii) If 2x - 3y = 10 and xy = 16; find the value of 8x³ - 27y³.
(iii) If x > 0 and x² + 1÷9x² = 25/36, find: x³ + 1÷27x³

Answer 1

given
     x = y + 2
       x² + y² = 34

     x - y = 2
   squaring,  (x - y)² = 2²
             x² + y² - 2 x y = 4
             34 - 2 x y = 4
               x y = 15
===========================
  2x - 3y = 10          x y  = 16

  8x³ - 27 y³ = (2x - 3y) ( 4x² + 6 xy + 9 y²)
               = (2 x - 3y) [ (2x -3y)² + 18 xy ]
               = 10 * [ 10² + 18 * 16 ]
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  x > 0  and  x² + 1/9x²  = 25 / 36
      (x + 1/3x)²  = x² + 1/9x² + 2/3  = 25/36  + 2/3
             (x + 1/3x)² = 49/36
                 x + 1/3x = 7/6      , we drop the negative value as x > 0.

   x³ +  1/27x³  = (x +  1/3x) ( x² + 1/9x² - x/3x)
                     = 7/6 * (25/36 - 1/3)
                     = 7/6 * 1/36
                      = 7/216

Answer 2

x=y+2 x^2+y^2=34 Put x=y+2 (y+2)^2 + y^2=34 y^2+4+4y+y^2=34 2y^2+4y+4=34 y^2+2y+2=17 y^2+2y-15=0 y^2+5y-3y-15=0 (y-3)(y+5)=0 Y=3, Y=-5 X=5,Y=3 Thus,xy=15.