In an exercise asking to mark true or false, it shows:
$$\frac{1}{a/x-b/x}=\frac{1}{a-b}$$
It really look like false to me. But the answer is true! How can it be?
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2$\begingroup$ which books is this from? $\endgroup$– shuhaloCommented Feb 24, 2011 at 2:37
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3 Answers
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Suppose $x\neq 0$, and $a\neq b$. Multiplying top and bottom of the left hand side by $x$ shows $$\frac{1}{a/x-b/x}=\frac{x}{a-b}$$ and this equals $\frac{1}{a-b}$ if and only if $x=1$.
In short, it can't be true, but my guess is that the book meant to have an $x$ as the numerator of the right hand side of the original equation.
answered Feb 23, 2011 at 23:51
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$\begingroup$ it really looks like the book's mistake.. $\endgroup$ Commented Feb 23, 2011 at 23:57
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2$\begingroup$ I edited to indicate that. Even books make mistakes from time to time. $\endgroup$ Commented Feb 24, 2011 at 0:00
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1$\begingroup$ One could also point out that they are not strictly equal-the RHS is defined at x=0, while the LHS is not $\endgroup$ Commented Feb 24, 2011 at 14:42
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Are you sure it is not $\frac{1}{a/x-b/x}=\frac{x}{a-b}$? Otherwise I agree with you.
answered Feb 23, 2011 at 23:49
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$\frac{1}{\frac{a}{x}-\frac{b}{x}} = \frac{1}{\frac{a-b}{x}} = \frac{x}{a-b} \not\equiv \frac{1}{a-b}$
So this is not true in the general case.
