I know this might seem very simple, but I can't seem to isolate x.
$$\frac{1}{x} = \frac{1}{a} + \frac{1}{b} $$
Please show me the steps to solving it.
$\endgroup$3 Answers
$\begingroup$$\frac{1}{x} = \frac{b}{ab} + \frac{a}{ab}$
$\frac{1}{x} = \frac{a + b}{ab}$
$x = \frac{ab}{a + b}$
note that $\frac{1}{x} = \frac{1}{a} + \frac{1}{b}$ is possible if and only if $\frac{1}{a} + \frac{1}{b} \neq 0$. This implies that $a \neq -b$; and, hence $a + b \neq 0$.
$\endgroup$ $\begingroup$You should combine $\frac1a$ and $\frac1b$ into a single fraction using a common denominator as usual:
$$\begin{eqnarray} \frac1x& = &\frac1a + \frac1b \\ &=&{b\over ab} + {a\over ab} \\ &=& b+a\over ab \end{eqnarray}$$
So we get: $$x = {ab\over{b+a}}.$$
Okay?
$\endgroup$ 3 $\begingroup$1/x = (a+b)/ab , x~=0
x=ab/(a+b),x~=0
