SOLUTION: What are all ordered pairs of real numbers (x, y) for which 17x^2 –10xy + 2y^2 –6x + 2 =0?

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Question 520709: What are all ordered pairs of real numbers (x, y) for which
17x^2 –10xy + 2y^2 –6x + 2 =0?

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
17x² – 10xy + 2y² – 6x + 2 = 0



Rearrange the terms in descending order of y



2y² - 10xy + 17x² - 6x + 2 = 0



Think of it like this:



2y² + (-10x)y + (17x² - 6x + 2) = 0 



and use the quadratic formula



with a = 2, b = -10x and c = 17x² - 6x + 2





y = %28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+



y = %28-%28-10x%29+%2B-+sqrt%28+%28-10x%29%5E2-4%2A%282%29%2A%2817x%5E2-6x%2B2%29+%29%29%2F%282%2A%282%29%29+



y = %2810x+%2B-+sqrt%28100x%5E2-8%2817x%5E2-6x%2B2%29+%29%29%2F4+



y = %2810x+%2B-+sqrt%28100x%5E2-136x%5E2%2B48x-16+%29%29%2F4+



y = %2810x+%2B-+sqrt%28-36x%5E2%2B48x-16+%29%29%2F4+



y = %2810x+%2B-+sqrt%28-4%289x%5E2-12x%2B4%29+%29%29%2F4+



y = %2810x+%2B-+sqrt%28-4%283x-2%29%283x-2%29+%29%29%2F4+



y = %2810x+%2B-+sqrt%28-4%283x-2%29%5E2%29%29%2F4+ 

y = %2810x+%2B-+2i%2Asqrt%28%283x-2%29%5E2%29%29%2F4+ 

Divide every term in the top and bottom by 2

y = %285x+%2B-+i%2Asqrt%28%283x-2%29%5E2%29%29%2F2+

y = %285x+%2B-+i%2Aabs%283x-2%29%29%2F2+


That will be imaginary unless the imaginary part is 0, so

3x-2 = 0

   x = 2%2F3

Substitute that

y = %285%282%2F3%29%29%2F2+ = %2810%2F3%29%2F2 = 10%2F3÷2 = 5%2F3

and get y = 5%2F3

Answer:  (2%2F3,5%2F3)

Edwin