How do you solve $\frac{3}{x - 3} + \frac{4}{x} + \frac{1}{3} = \frac{x}{x - 3}$ $3/(x-3) + 4/x +1/3 = x/(x-3)$ and find any extraneous solutions?

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Answer 1 · 2018-06-01T10:26:26

$x = 6$ $x=6$

$\frac{3}{x - 3} + \frac{4}{x} + \frac{1}{3} = \frac{x}{x - 3}$ $3/(x-3)+4/x+1/3=x/(x-3)$

$\frac{4}{x} + \frac{1}{3} = \frac{x - 3}{x - 3}$ $4/x+1/3=(x-3)/(x-3)$

I assumed $x \neq 3$ $x!=3$ , because $x = 3$ $x=3$ is extraneous solution. Hence,

$\frac{4}{x} + \frac{1}{3} = 1$ $4/x+1/3=1$

$\frac{4}{x} = \frac{2}{3}$ $4/x=2/3$

$x = 3 \cdot \frac{4}{2} = 6$ $x=3*4/2=6$

How do you solve 3x−3+4x+13=xx−3 3/(x-3) + 4/x +1/3 = x/(x-3) and find any extraneous solutions?