Solve for x1-p-1-q-1-x-1-x-p-qA- undefinedB- undefinedC- undefinedD- undefined

The given equation 1/p+1/q+1/x=1/p+q+x 1/p+1/q=1/p+q+x 1/x p+q/pq= x (p+q+x)/x(p+q+x) p+q/pq = (p+q)/x(p+q+x) Dividing by p + q on the both sides of the equ ...

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Solve for x1/...

Question

Solve for x
$(\frac{1}{p}) + (\frac{1}{q}) + (\frac{1}{x}) = \frac{1}{(x + p + q)}$

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Solution

The given equation
$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$
$\Rightarrow \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$
$\Rightarrow \frac{p + q}{p q} = \frac{x - (p + q + x)}{x (p + q + x)}$
$\Rightarrow \frac{p + q}{p q} = \frac{- (p + q)}{x (p + q + x)}$
Dividing by $p + q$ on the both sides of the equation:
$\Rightarrow \frac{1}{p q} = \frac{- 1}{x (p + q + x)}$
$\Rightarrow x (p + q + x) = - p q$
$\Rightarrow x^{2} + p x + q x + p q = 0$
$\Rightarrow x (x + p) + q (x + p) = 0$
$\Rightarrow (x + p) (x + q) = 0$
therefore
$x = - p$ or $x = - q$

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Solve for x (1p)+(1q)+(1x)=1(x+p+q)

The given equation 1p+1q+1x=1p+q+x ⇒1p+1q=1p+q+x−1x ⇒p+qpq=x−(p+q+x)x(p+q+x) ⇒p+qpq=−(p+q)x(p+q+x) Dividing by p+q on the both sides of the equation: ⇒1pq=−1x(p+q+x) ⇒x(p+q+x)=−pq ⇒x2+px+qx+pq=0 ⇒x(x+p)+q(x+p)=0 ⇒(x+p)(x+q)=0 thereforex=−p or x=−q

Solve for x
$(\frac{1}{p}) + (\frac{1}{q}) + (\frac{1}{x}) = \frac{1}{(x + p + q)}$