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x= 4-x/y^2-x 

 

How many ordered pairs of positive integers (x,y) satisfy the equation above. 

 

Multiple Choice answers

 

a) 0

b) 1

c) 2

d) 3

e) 4

Guest Jan 2, 2018
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1+0 Answers

 #1
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x =  (4 - x) / (y^2 - x)

 

y^2x  - x^2  =   4 - x

 

y^2x  =   x^2 - x  +   4

 

y^2  =   [ x^2  -  x  +  4 ]  /  x

 

y^2  =  x  -  1   +  4/x

 

Notice  that  the first two terms always result in an integer for any integer value of x

 

But   4/x     is only an integer  when x  =  ±1, ±2 or ±4

 

But we can reject  the negative values  because  they make the right side negative.....so y is not a real number  for these values

 

And when x = 2,  y  = ±√3    which isn't an integer pair

 

So.....when  x  =  1,  y   = ±2

And when  x  =   4,  y   = ±2

 

So......the ordered pairs of positive integers that make this true  are  (1,2), (-1,2), (4, 2)  and (4 , -2)

 

So....the correct answer is  (e)

 

 

 

 

cool cool cool

CPhill  Jan 2, 2018
edited by CPhill  Jan 3, 2018

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