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 ...

224

You visited us 224 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Quantitative Aptitude

Equations

Solve for x1/...

Question

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

Open in App

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 $x = - q$

Suggest Corrections

Similar questions

s i n {2 t a n^{- 1} \sqrt{\frac{1 - x}{1 + x}}}

Q. Evaluate:

l i m_{x \to 1} [\frac{1}{x^{2} + x - 2} - \frac{x}{x^{2} - 1}]

Q. Find the values of p and q, if the polynomial

x^{4} + p x^{3} + 2 x^{2} - 3 x + q

is divisible by the polynomial

x^{2} - 1.

Q. Solve the equation

9 x^{2} - 155 x - 500

using quadratic method.

Q. Quardratic surds
If

x = \sqrt{\frac{\sqrt{5} + 1}{\sqrt{5} + 1}}

Then prove that

x^{2} - x - 1 = 0

Join BYJU'S Learning Program

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