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$\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4} < 0\right\}$ is equal to
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Verified Answer
The correct answer is:
$(-1,1) \cup(4,6)$
$\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4} < 0\right\}$
$\frac{14 x(x-4)-(9 x-30)(x+1)}{(x+1)(x-4)} < 0$
$\frac{14 x^2-56 x-\left(9 x^2-30 x+9 x-30\right)}{(x+1)(x-4)} < 0$
$\frac{\left(14 x^2-56 x-9 x^2+30 x-9 x+30\right)}{(x+1)(x-4)} < 0$
$\frac{\left(5 x^2-35 x+30\right)}{(x+1)(x-4)} < 0$
$\frac{\left(x^2-7 x+6\right)}{(x+1)(x-4)} < 0, \frac{(x-1)(x-6)}{(x+1)(x-4)} < 0$
Drawn number line,
Hence,
$x \in(-1,1) \cup(4,6)$
$\frac{14 x(x-4)-(9 x-30)(x+1)}{(x+1)(x-4)} < 0$
$\frac{14 x^2-56 x-\left(9 x^2-30 x+9 x-30\right)}{(x+1)(x-4)} < 0$
$\frac{\left(14 x^2-56 x-9 x^2+30 x-9 x+30\right)}{(x+1)(x-4)} < 0$
$\frac{\left(5 x^2-35 x+30\right)}{(x+1)(x-4)} < 0$
$\frac{\left(x^2-7 x+6\right)}{(x+1)(x-4)} < 0, \frac{(x-1)(x-6)}{(x+1)(x-4)} < 0$
Drawn number line,
Hence,
$x \in(-1,1) \cup(4,6)$
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