Question

In the attachment is a solution of inequality problem. There is a mistake in my answer. Find it out and correct it. Also give full solution of the question.
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Answer 1

I hope this is correct

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Answer 2

$\large\underline{\sf{Given \:Question - }}$

Solve the inequality :-

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 19x + 88}{ {x}^{2} - 11x + 24} \leqslant 0$

$\green{\large\underline{\sf{Solution-}}}$

Given inequality is

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 19x + 88}{ {x}^{2} - 11x + 24} \leqslant 0$

By splitting of middle terms, we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 11x - 8x + 88}{ {x}^{2} - 8x - 3x + 24} \leqslant 0$

$\rm :\longmapsto\:\dfrac{x(x - 11) - 8(x - 11)}{x(x - 8) - 3(x - 8)} \leqslant 0$

$\rm :\longmapsto\:\dfrac{(x - 11)(x - 8)}{(x - 8)(x - 3)} \leqslant 0$

Let now, first we define the domain.

$\rm :\longmapsto\:\dfrac{(x - 11)(x - 8)}{(x - 8)(x - 3)} \leqslant 0 \: \: provided \: that \: x \: \ne \: 3,8$

$\rm :\longmapsto\:\dfrac{x - 11}{x - 3} \leqslant 0 \: \: provided \: that \: x \: \ne \: 3,8$

$\rm :\longmapsto\:\dfrac{(x - 11)(x - 3)}{(x - 3) {}^{2} } \leqslant 0 \: \: provided \: that \: x \: \ne \: 3,8$

$\rm :\longmapsto\:(x - 11)(x - 3) \leqslant 0 \: \: provided \: that \: x \: \ne \: 3,8$

$\bf\implies \:x \: \in \: (3,8) \: \cup \: (8,11]$

Additional Information :-

If a and b are real numbers such that a < b, then

$\boxed{ \bf{ \: (x - a)(x - b) < 0\bf\implies \:x \in \: (a,b)}}$

$\boxed{ \bf{ \: (x - a)(x - b) \leqslant 0\bf\implies \:x \in \: [a,b]}}$

$\boxed{ \bf{ \: (x - a)(x - b) \geqslant 0\bf\implies \:x \in \: ( - \infty ,a] \cup \: [b, \infty )}}$

$\boxed{ \bf{ \: (x - a)(x - b) > 0\bf\implies \:x \in \: ( - \infty ,a) \cup \: (b, \infty )}}$

$\boxed{ \bf{ \: \frac{x - a}{x - b} > 0\bf\implies \:x \in \: ( - \infty ,a) \cup \: (b, \infty )}}$

$\boxed{ \bf{ \: \frac{x - a}{x - b} \geqslant 0\bf\implies \:x \in \: ( - \infty ,a] \cup \: (b, \infty )}}$

$\boxed{ \bf{ \: \frac{x - a}{x - b} > 0\bf\implies \:x \in \: ( - \infty ,a) \cup \: (b, \infty )}}$