x + y + xy = 19
y + z + yz = 29
z + x + zx = 5
If x, y, and z are positive real numbers satisfying the system above, then find x, y, and z.
+129899 x + y + xy = 19 (1)
y + z + yz = 29 (2)
z + x + zx = 5 (3)
Simplify (1) Simplify (3)
y ( x + 1) + x = 19 z ( x + 1) + x = 5
y ( x + 1) = 19 - x z ( x + 1) = 5 - x
y = (19- x) / ( x + 1) z = ( 5 - x) / ( x + 1)
Sub these into (2) for y,z
(19-x) / ( x + 1) + (5 -x) / (x + 1)) + ( 19-x) ( 5 - x) / [ (x + 1)^2 ] = 29 simplify
Multiply through by ( x + 1)^2
(19 - x) (x + 1) + ( 5 - x) (x + 1) + (19 - x) (5 -x) = 29 [ (x + 1) ] ^2
-x^2 + 18x + 19 + -x^2 + 4x + 5 + 95 - 24x + x^2 = 29 [ ( x + 1)^2 ]
-x^2 - 2x + 119 = 29x^2 + 58x + 29
30x^2 + 60x - 90 = 0 divide through by 30
x^2 + 2x - 3 = 0 factor
(x + 3) ( x - 1) = 0 since x,y,z are positive then
x = 1
And
z = (5-1) / ( 1 + 1) = 4 / 2 = 2
And
y = (19-1) / ( 1 + 1) = 18 / 2 = 9
