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If the replacement set is the set of natural numbers, solve: 3x – 4 > 6 |
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Answer» 5 < 0 X – 5 + 5 < 0 + 5 ………(Adding 5) => x < 5 Required ANSWER = {1, 2, 3, 4} (II) x + 1 ≤ 7 = > x + 1 – 1 ≤ 7 – 1 (Subtracting 1) => x ≤ 6 Required answer = {1, 2, 3, 4, 5, 6} (iii) 3x – 4 > 6 3x – 4 + 4 > 6 + 4 (Adding 4) => 3x > 10 \frac { 3x }{ 3 } > \frac { 10 }{ 3 } …(DIVIDING by 3) => x > \frac { 10 }{ 3 } => x > 3\frac { 1 }{ 3 } Required answer = { 4, 5, 6, …} (iv) 4x + 1 ≥ 17 => 4x + 1 – 1 ≥ 17 – 1 (Subtracting) => 4x ≥ 16 => \frac { 4x }{ 4 } ≥ \frac { 16 }{ 4 } (Dividing by 4) => x ≥ 4 Required answer = {4, 5, 6, …} |
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