I have an example rational equation in a textbook which I was able to solve:
$$\dfrac{3}{x-6} = \dfrac{5}{x}$$
Least common denominator is : $x(x-6)$ so:
$$\frac{x(x-6)3}{x-6} = \dfrac{x(x-6)5}{x} = 3x = 5(x - 6)$$
$x = 15$
However, in my textbook, in addition I'm asked to provide "the excluded values". The provided solution says that the excluded values here are $6$ and $0$.
What does that mean and why are they $6$ and $0$?
$\endgroup$ 12 Answers
$\begingroup$You may not divide with $0$ so $x-6\ne0$ and $x\ne 0$. And if you eventually get one of those as a solution you must refute it. Say you have $${x^2-1\over x-1}=2$$ after canceling we get $x+1=2$ so $x=1$. But this value is not OK since we would divide by $0$ at start.
$\endgroup$ $\begingroup$The excluded values of $x$ are those values, when substituted into any rational expression, that would make the denominator equal to $ 0 $. Any rational expression with the denominator being $ 0 $ is not defined.
In this case, the values of $x$ are $0$ and $6$ because $$\frac{3}{6-6} = \frac{3}{0} = undefined $$ and $$\frac{5}{0} = undefined $$
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