x+y=9 and x+y=27; x and y ?

[Guest]

Please write the solution not only result.
[tex]\sqrt{x}+y=9[/tex] and [tex]x+\sqrt{y}=27[/tex]
Find x = ? and y = ?

[Guest]

Let
[tex]a = \sqrt{x}[/tex],
[tex]b = \sqrt{y}[/tex]
to save typing the square root symbol over and over.

So the problem is
[tex]a+b^2 = 9[/tex]
[tex]a^2+b = 27[/tex]
with [tex]a\geq 0[/tex] and [tex]b\geq 0[/tex] (because we are taking the positive square roots).

If we assume there is an integer solution for [tex]a[/tex] and [tex]b[/tex], there are only a few cases to check:
[tex]b = 0, 1, 2, 3[/tex]
(higher values of [tex]b[/tex] cause [tex]a[/tex] to be negative).
It is easy to check only [tex]b=2[/tex] works (with [tex]a=5[/tex]). This corresponds to [tex]x=25[/tex], [tex]y=4[/tex].

There are no other solutions, to see why observe that the first equation tells us that [tex]a = 9-b^2[/tex], and substituting it into the second gives
[tex](9-b^2)^2+b = 27[/tex]
which rearranges to
[tex]b^4-18b^2+b+54 = 0[/tex].
We know [tex]b=2[/tex] is a solution which means [tex](b-2)[/tex] is a factor of the polynomial
[tex]b^4-18b^2+b+54 = (b-2)(b^3+2b^2-14b-27)[/tex]
So any other solutions satisfy
[tex]b^3+2b^2-14b-27 = 0[/tex].
A graph of the polynomial [tex]b^3+2b^2-14b-27[/tex] would look like a peak followed by a trough, or put another way the value increases from -infinity then decreases for a bit then increases again to +infinity.
At [tex]b=-3[/tex] the polynomial takes a value of [tex]6[/tex].
At [tex]b=0[/tex] the polynomial takes a value of [tex]-27[/tex].
So the polynomial is decreasing at [tex]b=0[/tex] or is in the second increasing phase (we can't still be in the first increasing phase as the value has decreased).
At [tex]b=3[/tex] the polynomial is [tex]-24[/tex] which is still negative. So there is no root between [tex]0[/tex] and [tex]3[/tex] (if there was a root the polynomial would have had to "increase then decrease", but we know it is either "increasing" or "decreasing then increasing").

Out of interest you can find the other solutions of the cubic but it involves using trigonometric functions and the answers aren't pretty.

Hope this helped,

R. Baber.