# Solving a literal equation containing fractions.

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.

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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?

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and x~=0. Correct ! –  Zeta.Investigator Aug 20 '12 at 17:31
How exactly did you flip ${1\over{x}} = {b + a\over{ab}}$ to $x = {ab\over{b+a}}$ ? –  Daniel Pendergast Aug 20 '12 at 17:34
@Dan: Suppose that $\frac{u}v=\frac{x}y$. Multiply through by $vy$ to get $uy=xv$, then divide through by $ux$ to get $\frac{y}x=\frac{v}u$. Alternatively, multiply both sides of the original equation by $\frac{y}x$ t0 get $\frac{uy}{vx}=1$, then multiply both sides of that equation by $\frac{v}u$ to get $\frac{y}x=\frac{v}u$. Whenever two non-zero fractions are equal, their reciprocals (obtained by turning them upside down) are also equal. –  Brian M. Scott Aug 20 '12 at 17:41
Awesome. Thank you. –  Daniel Pendergast Aug 20 '12 at 17:43

$\frac{1}{x} = \frac{b}{ab} + \frac{a}{ab}$

$\frac{1}{x} = \frac{a + b}{ab}$

$x = \frac{ab}{a + b}$

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 $a + b \neq 0$.

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1/x = (a+b)/ab , x~=0
x=ab/(a+b),x~=0

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