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Introduction to linear inequalities

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Question

$Number of elements in the set T = \{x ∣ \frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x}\} i s$

Moderate

Solution

$We have, T = \{x ∣ \frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x}\}$
$\begin{array}{l} ∴ \frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x} \\ \Rightarrow \frac{x + 5 - 5 (x - 7)}{x - 7} = \frac{4 x - 40}{13 - x} \\ \Rightarrow \frac{x + 5 - 5 x + 35}{x - 7} = \frac{4 x - 40}{13 - x} \\ \Rightarrow \frac{- 4 x + 40}{x - 7} = \frac{4 x - 40}{13 - x} \\ \Rightarrow - (4 x - 40) (13 - x) = (4 x - 40) (x - 7) \\ \Rightarrow (4 x - 40) (x - 7 + 13 - x) = 0 \\ \Rightarrow x - 10 = 0 \\ \Rightarrow x = 10 \\ \Rightarrow T = {10} \end{array}$
$Hence, number of elements in T is =1$

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The solution of $x - 1 = (x - [x]) (x - {x})$ (where $[x]$ and ${x}$ are the integral and fractional part of $x$ ) is :

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