3
$\begingroup$

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?

  • 2
    which books is this from?2011-02-24

3 Answers 3

3

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.

  • 1
    One could also point out that they are not strictly equal-the RHS is defined at x=0, while the LHS is not2011-02-24
1

Are you sure it is not $\frac{1}{a/x-b/x}=\frac{x}{a-b}$? Otherwise I agree with you.

1

$\frac{1}{\frac{a}{x}-\frac{b}{x}} = \frac{1}{\frac{a-b}{x}} = \frac{x}{a-b} \not\equiv \frac{1}{a-b}$

So this is not true in the general case.