Solve: $5 x - 3 < 3 x + 1$ , where $x$ is an integer.

{- 3, - 2, - 1, 0, 1, 2, 3}

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{- 5, - 4, - 3, - 2, - 1, 0, 1, 2}

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{- 4, - 3, - 2, - 1, 0, 1}

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{- 2, - 1, 0, 1, 2}

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Solution

The correct option is B ${- 4, - 3, - 2, - 1, 0, 1}$
Given, $5 x - 3 < 3 x + 1$

$⟹ 5 x - 3 x < 1 + 3$

$⟹ 2 x < 4 ⟹ x < 2$

From the options, option C has the set with values $< 2$

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The solution set for the inequation -3 $\leq$ x $\leq$ 3, x $\in$ I is

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