How many ordered pairs of real numbers ( x , y ) satisfy: ( x − 1 ) ( y 2 + 6 ) = y ( x 2 + 1 ) , ( y − 1 ) ( x 2 + 6 ) = x ( y 2 + 1 ) ?
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Notice that the two relations are inverses of each other since each y in the first relation is interchanged for an x in the second and each x in the first relation is interchanged for a y in the second equation. Inverse relations only intersect on the line of symmetry y=x.
This means we can substitute each y for an x in one of the equations. When we simplify we get a parabola with a positive discriminant in the variable x giving us two solutions. The same can be done if we replace each x for a y. The process is identical to the last and gives us two more different roots. Thus we have 4 solutions to the system of equations.
Could you explain a little bit your method?
Note that ( 2 , 3 ) is a solution to the problem, which contradicts your claim.
Using your approach, the only solutions to
y = − x x = − y
would be when x = y , hence x = y = 0 . But we know that there are many other solutions with x = y .
cyclic order (x-1)(x^2 + 6 ) = x(x^2 + 1 ) solve x = 2 , 3 x=2 in equation1 solve y = 2 , 3 x=3 in equation1 solve y = 2 , 3 ( x , y ) = ( 2 , 2 ) , ( 2 , 3 ) ( 3 , 2 ) , ( 3 , 3 ) Ans = 4 pairs.
You should explain what you meant by "cyclic order (x-1)(x^2 + 6 ) = x(x^2 + 1 )". It looks unclear.
As I understand, cyclic order means replace x with y AND y with x. You have only replaced y with x. Please explain the theory behind your method.
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If you sum the two equations you get the equation of a circle, while if you subtract them you get ( x − y ) times the equation of an hyperbola. Simple algebra follows.