Question

$\sf{If \: \bold{px + qx - 12pq = 0} , find \: the \: value \: of:}$
$\\ \rightarrow{\tt{ \dfrac{x + 6p}{x - 6p} + \dfrac{x + 6q}{x - 6q}}}$

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Answer 1

[x] 2 −5[x]+6≤0

⇒([x]−2)([x]−3)≤0

⇒[x]∈[2,3]

∴x∈[2,4)

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Answer 2

I'm not sure about this method but I'll try.

We know that

$\red{\bigstar}px+qx-12pq=0$

then,

$\red{\bigstar}(p+q)x=12pq$

$\text{(Given value)}$

$=\dfrac{x+6p}{x-6p}+\dfrac{x+6q}{x-6q}$

(multiplying 1 on both fractions, we make both fractions contain $(p+q)x$)

$=\dfrac{p+q}{p+q}\times\left(\dfrac{x+6p}{x-6p}+\dfrac{x+6q}{x-6q}\right)$

$=\dfrac{(p+q)x+6p(p+q)}{(p+q)x-6p(p+q)}+\dfrac{(p+q)x+6q(p+q)}{(p+q)x-6q(p+q)}$

(taking substitution of $(p+q)x=12pq$)

$=\dfrac{12pq+6p(p+q)}{12pq-6p(p+q)}+\dfrac{12pq+6q(p+q)}{12pq-6q(p+q)}$

(the numerator and denominator of each fraction has HCF of $6p$ and $6q$ respectively)

$=\dfrac{2q+(p+q)}{2q-(p+q)}+\dfrac{2p+(p+q)}{2p-(p+q)}$

$=\dfrac{p+3q}{-p+q}+\dfrac{3p+q}{p-q}$

$=\dfrac{-(p+3q)+(3p+q)}{p-q}$

$=\dfrac{2p-2q}{p-q}$

$=2$