How many are there?
Algebra
Level
4
How many real numbered ordered pairs satisfy the following inequalities:
more than 1, but finitely many
infinitely many
0
1
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1 solution
Let's see if rewriting the inequalities will work.
For the first inequality, ( x + 2 y ) 2 ≤ ( 2 x − 3 ) ( 4 y + 3 ) x 2 + 4 x y + 4 y 2 ≤ 8 x y + 6 x − 1 2 y − 9 x 2 + 4 y 2 + 9 − 4 x y − 6 x + 1 2 y ≤ 0 ( x − 2 y − 3 ) 2 ≤ 0 But we know that all squares are nonnegative, that is, ( x − 2 y − 3 ) 2 ≥ 0 . Therefore, x − 2 y − 3 = 0 .
Similarly, for the second inequality, ( x − 2 y ) 2 ≤ ( 5 − 2 x ) ( 4 y − 5 ) x 2 − 4 x y + 4 y 2 ≤ 2 0 y − 2 5 − 8 x y + 1 0 x x 2 + 4 y 2 + 2 5 + 4 x y − 1 0 x − 2 0 y ≤ 0 ( x + 2 y − 5 ) 2 ≤ 0 Since ( x + 2 y − 5 ) 2 ≥ 0 for all x , y , thus x + 2 y − 5 = 0 .
Adding the two resulting equations (and simplifying) gives x = 4 . Thus, form the first equation, y = 2 1 . Thus, ( 4 , 2 1 ) is the unique ordered pair solution. The answer is 1 .