Find the solution set and its nature of the following equations and(xbox,y)€z.
i.)x+y=8
ii.)x-y=0
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What is the number of integral solutions (x,y) of the system of equations x² - xy + 8 = 0 and x² - 8 x + y=0?
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x^2 - xy + 8 = 0 ……..(1)
x^2 - 8x + y = 0 => y = 8x - x^2 ………(2)
Using (2) in (1) we get x^2 - x(8x - x^2) + 8 = 0
=> x^2 -8x^2+x^3 +8 = 0
=> x^3 - 7x^2 + 8 = 0
Since p(-1) = 0 if we take p(x) as x^3–7x^2+8, x+1 is a factor of x^3–7x^2 + 8.
So x^3 - 7x^2 + 8 = (x+1)(x^2 + kx + 8)
Comparing x^2 term we get -7 = k+1 => k = -7–1
=> k = -8 . So x^3 - 7x^2 + 8 = 0
=> (x+1)(x^2 - 8x + 8)= 0
=> x = -1 or x = {8 + or - (64–32)}/2 = 4 + or - 2√2
So x = -1 is the only integral solution.