Question #58346

1 Answer
Sep 17, 2016

Answer:

#x=9/2+sqrt17/2, y=-9/2+sqrt17/2#,

and,

#x=9/2-sqrt17/2, y=-9/2-sqrt17/2#.

Explanation:

Let #(x-y)^(1/2)=a :. (x-y)=a^2#. Using these in the #1^(st)# eqn., we

get, #3a^2+a-30=0 rArr a=-10/3,or, a=3#

#:. (x-y)^(1/2)=-10/3" is not admissible, since, "(x-y)^(1/2) >0.#

#:. (x-y)^(1/2)=3 rArr x-y=9, or, x=y+9#.

By the #2^(nd)# eqn., then,

#(y+9)y+27=11, or, y^2+9y+16=0#

#:. y=(-9+-sqrt(81-64))/2=(-9+-sqrt17)/2=-9/2+-sqrt17/2#.

Corresponding #x=y+9=-9/2+-sqrt17/2+9=9/2+-sqrt17/2#.

Thus, #x=9/2+sqrt17/2, y=-9/2+sqrt17/2#, and,

#x=9/2-sqrt17/2, y=-9/2-sqrt17/2#.

These roots satisfy the eqns. Hence, the soln.