Let x=f(u,v) and y=g(u,v), then xy=9⇒fg=9.

Let f=u+av+3 and g=u+bv+3, then fg=u^2+buv+3u+auv+abv^2+3av+3u+3bv+9=9.

[The mappings can also be expressed: f:u+av+3→x and g:u+bv+3→y.

Also, the constant 3 for both mappings has been arbitrarily selected because 3*3=9. Other constants, p and q, could have been chosen where pq=9.]

If a=-b then the uv term disappears and u^2+3u-a^2v^2+3u=0.

If a=1, **u^2+6u-v^2=0**; f=x=u+v+3 and g=y=u-v+3; v=(x-y)/2; u=(x+y-6)/2.

The general form of the answer is u^2+6u-a^2v^2=0, where a is a constant, using 3 as the arbitrary constant.

[A more general solution would be: u^2-a^2v^2+u(p+q)+av(q-p)=0 where pq=9.]

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