**problem**: If c is a real number with 0 < c < 1, then show that the values taken by the function y = \({frac {x^2+2x+c}{x^2+4x+3c}} \) , as x varies over real numbers, range over all real numbers.

**solution:** y = \({\frac {x^2+2x+c}{x^2+4x+3c}} \)

or \({yx^2} \) + 4yx + 3cy = \({x^2} \) + 2x + c

or \({x^2} \) (y-1) + (4y-2)x + 3cy – c = 0

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