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Modulus Function

Solve 1a-b+...

Question

Solve $\frac{1}{a} - b + x = \frac{1}{x} + \frac{1}{a} - \frac{1}{b}$ where $a \neq= 0$ , $b \neq 0$ $x \neq 0$ ( $x = - a, x = b)$

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Solution

$\frac{1}{a} - b + x = \frac{1}{x} + \frac{1}{a} - \frac{1}{5}$
$\frac{1}{a} - b + x - \frac{1}{a} = \frac{1}{x} - \frac{1}{b}$
$\Rightarrow \frac{x - b}{1} = \frac{b - x}{x b}$
$\Rightarrow \frac{x - b}{1 (b - x)} = \frac{1}{x b}$
$\Rightarrow x b = \frac{- (x - b)}{(x - b)}$
$\Rightarrow x = \frac{- 1}{b}$

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x : \frac{1}{(a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}

{a \neq 0, b \neq 0, x \neq 0, x \neq - (a + b)}

x : \frac{1}{(a a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, [a \neq 0, b \neq 0, x \neq 0, x \neq - (a + b)]

Q. Solve for x :
1/a+b+x = 1/a +1/b +1/x (a≠0, b≠0, x≠0).

x

\frac{1}{a} + \frac{1}{b} + \frac{1}{x} = \frac{1}{a + b + x};

a \neq 0, b \neq 0, x \neq 0.

x : \frac{1}{(a + b + x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, [a \neq 0, b \neq 0, x \neq 0, x \neq - (a + b)]

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