How many answers are there?
( X + Y ) ( X + Y ) ( Y + Z ) ( Y + Z ) ( X + Z ) ( X + Z ) = = = 4 Z − 1 4 X − 1 4 Y − 1
If X , Y and Z are rational real numbers that satisfy the system of equations above, find the value of X + Y + Z .
The answer is 1.5.
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1 solution
The equations are cyclic so x=y=z
I did some calculations can you tell what should I do now ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 + ( x + y + z ) 2 = 2 ( x + y + z )
We can show that x=y=z is the only solution.
Proof by contradiction: suppose x=y=z is not satisfied, particularly x=y is not true. Set z as the subject in equation 1, then substitute it into equation 2 and 3, then equation 2 - equation 3 to get equation 4. Divide equation 4 by (x-y) because x = y . So you get a quadratic equation of x + y . The discriminant of that equation is negative, so there's no real solution. Thus x=y=z is the only solution.
I haven't got a full solution, but I know that X=Y=Z produces X = 0.5 such that X+Y+Z = 1.5. Maybe an inequality is needed to show that this condition must apply.