Solutions of the system of inequalities is 3x-4 5x-4 11x-9- 15x-3

The correct option is A 2Given, 3x4 lt;8x+6 = gt; 5x lt; 10 = gt; x lt; 2 = gt; x gt; 2 nbsp; nbsp; — (1 ...

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Question

Solutions of the system of inequalities is $3 x - 4 < 8 x + 6; 2 x - 1 > 5 x - 4$ & $11 x - 9 \leq 15 x + 3$

- 2 < x < 1

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- 4 < x < 1

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- 1 < x < 1

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- 2 < x < 2

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Solution

The correct option is A $- 2 < x < 1$
Given,
$3 x 4 < 8 x + 6$
$=> - 5 x < 10$
$=> - x < 2$
$=> x > - 2$ --- (1)
And $2 x 1 > 5 x 4$
$=> - 3 x > - 3$
$=> x < 1$ --- (2)
And $11 x 9 \leq 15 x + 3$
$=> - 4 x \leq 12$
$=> - x \leq 3$
$x \geq - 3$ --- (3)
From $(1), (2), (3)$ , we have $- 2 < x < 1$

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$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$