Ordered how?

Algebra Level 3

{ x + y − x − y = y x 2 − y 2 = 144 \large \begin{cases} { \sqrt{x+y} - \sqrt{x-y} = \sqrt y } \\ { x^2-y^2=144 } \end{cases}

Find the number of all ordered pairs ( x , y ) (x,y) that satisfies the equations above.


The answer is 2.

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Majed Musleh by Majed Musleh , 28, Palestine

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1 solution

Prakhar Gupta
Jun 14, 2015

The first equation is :- x + y − x − y = y \sqrt{x+y} - \sqrt{x-y} = \sqrt{y} Squaring both sides:- x + y + x − y − 2 x 2 − y 2 = y x+y+x-y -2\sqrt{x^{2}-y^{2}} = y From second equation:- 2 x − 2 144 = y 2x -2\sqrt{144} = y 2 x − 24 = y … ( 3 ) 2x-24 = y \ldots(3) Substituting ( 3 ) (3) in ( 2 (2 ):- x 2 − ( 2 x − 24 ) 2 = 144 x^{2} - (2x-24)^{2} = 144 3 x 2 − 96 x + 720 = 0 3x^{2} -96x+720 = 0 x 2 − 32 x + 240 = 0 x^{2} -32x+240=0 x = 20 , 12 x = 20,12 From ( 3 ) (3) :- y = 16 , 0 y = 16,0 We can see that both the equation satisfy the ordered pair ( 20 , 16 ) (20,16) and ( 12 , 0 ) (12,0) .

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Moderator note:

This is correct. There is no harm in checking back whether the solutions actually fit especially when you had square both sides of the equation. Nice standard approach.

Now if we included complex solutions there would be 2 solutions as 2i sqrt 3 - 2i sqrt 3 is also zero, x and y were not specified to be real numbers, but I got 2 real solutions and 2 complex solutions...but yes, 2 real solutions to this problem.

Joe Potillor - 4 years, 7 months ago

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