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Solve the following inequations: (a) `x + 5 lt 7, x in R` (b) `4x -3 ge 17, x in Z` (c) `3x -2 lt 1, x in N` |
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Answer» (a) Subtracting 5 from both sides, we get `x + 5 - 5 lt 7 - 5 rArr x lt 2` The set of all the number which are less than 2 is the solution set of the given inequation (b) Adding 3 on both sides, we get `4x - 3 + 3 ge 17 + 3` `4x ge 20` Dividing both sides by 4, we get `(4x)/(4) ge (20)/(4) rArr x ge 5` The set of all the integers which are greater than or equal to 5 is the solution set of the given inequation. (c) Adding 2 on both sides of `3x - 2 + 2 lt 1 + 2`, we get `3x lt 3` Dividing both sides by 3, we get `(3x)/(3) lt (3)/(3) rArr x lt (3)/(3) rArr x lt 1` There is no natural number which is less than 1 `:.` There is no solution for the given inequation in the set of natural numbers |
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