Chapter 15: Linear Inequations – Exercise 15.3
Linear Inequations – Exercise – 15.3 – Q.1
Consider the first inequation,
Consider the second inequation,
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Linear Inequations – Exercise – 15.3 – Q.2
We have,
|4 - x| + 1 - 3 < 0
⟹ |4 - x| - 2 < 0
⟹ 4 - x - 2 < 0
⟹ 2 - x < 0
⟹ - x < -2
⟹ x > 2 ....(ii)
Case II: When |4 - x| < 0
|4 - x| - 2 < 0
⟹ -(4 - x) -2 < 0
⟹ - 4 + x - 2 < 0
⟹ x - 6 < 0
⟹ x < 6 ....... (iii)
Combining (ii) and (iii) we get (2, 6) as the solution set.
Linear Inequations – Exercise – 15.3 – Q.3
Case I: When |3x - 4| ≥ 0
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⟹ 18x – 24 – 5 ≤ 0
⟹ 18x – 29 ≤ 0
⟹ 18x ≤ 29
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Case II: When |3x - 4|< 0
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Linear Inequations – Exercise – 15.3 – Q.4
We have,
Case I: When |x - 2| ≥ 0
x ≥ 2
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⟹ x – 2 ≥ 0
⟹ x ≥ 2 … (ii)
Case II: When |x - 2| < 0
x < 2
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⟹ - (x - 2) > 0
⟹ - x + 2 < 0
⟹ - x < - 2
⟹ x > 2 …… (iii)
Combining (ii) and (iii) we get (2, ∞) as the solution set.
Linear Inequations – Exercise – 15.3 – Q.5
We have,
Case I: When |x| ≥ 0 ⟹ x ≥ 0
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⟹ – x + 5 < 0
⟹ x > 5 ….. (ii)
Case II: When |x|< 0, x < 0
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⟹ 2 + x + 3 < 0
⟹ x + 5 < 0
⟹ x < - 5 …. (iii)
Combining (ii) and (iii) we get (-∞, -5) ∪ (-3, 3)∪(5, ∞) as the solution set.
Linear Inequations – Exercise – 15.3 – Q.6
We have,
Case I: When |x + 2| ≥ 0
i. e. x ≥ - 2
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⟹ - 2x + 2 < 0
⟹ - 2x < - 2 and x > 0
⟹ x > 1 …….. (ii)
Case II: |x + 2| < 0
i. e. x < - 2
⟹ - (x + 2) - 3x < 0
⟹ - x – 2 - 3x < 0
⟹ - 4x – 2 < 0
⟹ - 4x < 2
⟹ x > -1/2 …… (iii)
and x < 0
Combining (ii) and (iii) we get (-∞, 0) ∪ (1, ∞) as the solution set.
Linear Inequations – Exercise – 15.3 – Q.7
We have,
Case I: When |2x - 1| ≥ 0
i. e. 2x – 1 ≥ 0
2x ≥ 1
x ≥ 1/2
⟹|2x - 1| - 2x + 2 > 0 and x - 1 > 0
⟹ 2x – 1 - 2x + 2 > 0 and x > 1
⟹ x > 1 …….. (ii)
Case II: When |2x - 1|< 0
i. e. 2x – 1 < 0
2x < 1
x < 1/2
⟹ - (2x - 1) -2x + 2 > 0 and x < 1
⟹ - 4 + 3 > 0
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Linear Inequations – Exercise – 15.3 – Q.8
We have,
|x - 1| + |x - 2| + |x - 3| - 6 ≥ 0 ...... (i)
Case I: |x - 1| ≥ 0
x ≥ 1
⟹ x -1 -(x - 2) - (x - 3) - 6 ≥ 0
⟹ -x + 4 - 6 ≥ 0
⟹ -x ≥ 2
⟹ x ≤ -2
⟹ (-∞,-2] ..... (ii)
Case II: |x - 2| ≥ 0
x ≥ 2
⟹ x - 1 + x - 2 - (x - 3) - 6 ≥ 0
x - 6 ≥ 0
x ≥ 6
⟹ [6, ∞) ....... (iii)
Case III: When |x - 3| ≥ 0
x ≥ 3
⟹ x - 1 + x - 2 + x - 3 - 6 ≥ 0
⟹ 3x - 12 ≥ 0
⟹ 3x ≥ 12
⟹ x ≥ 4
⟹ ∴ x ϵ [4, ∞)
also
⟹ |x - 1| < 0
⟹ x < 1
⟹ -(x - 1) - (x - 2) - (x - 3) - 6 ≥ 0
⟹ -3x ≥ 0
⟹ x ≤ 0
⟹ |x - 2| < 0
x < 2
⟹ (x - 1) - (x - 2) - (x - 3) - 6 ≥ 0
⟹ x - 1 - x + 2 - x + 3 - 6 ≥ 0
⟹ -x -2 ≥ 0
⟹ x ≤ -2
⟹ |x - 3| < 0
⟹ x < 3
⟹ (x - 1) + (x - 2) - (x - 3) - 6 ≥ 0
⟹ x - 6 ≥ 0
⟹ x ≥ 6
Combining all cases we get (-∞, 0] ∪ [4, ∞) as the solution set.