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looi76 · #1 May 2nd, 2008, 06:35 AM

Question:
Find the value of $x$ when $y = 0$
$y = \frac{2x - 1}{x + 3}$

Attempt:

$(2x - 1)(x + 3)^{-1} = 0$

$(2x - 1)(x^{-1} + \frac{1}{3}) = 0$

$2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0$

$\frac{2}{3}x - x^{-1} + \frac{5}{3} = 0$

$\frac{4}{9}x^2 - 1 + \frac{5}{3} = 0$

$\frac{4}{9}x^2 + \frac{5}{3} - 1 = 0$

I think the quadratic equation that i got is wrong

... Where did I go wrong?