Example 16 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at Feb. 17, 2020 by Teachoo
Transcript
Example 16 Solve – 8 ≤ 5x – 3 < 7. –8 ≤ 5x – 3 < 7 Adding 3 all sides (Eliminating -3) – 8 + 3 ≤ 5x – 3 + 3 < 7 + 3 –5 ≤ 5x < 10 Dividing 5 all sides (Eliminating 5) (−5)/5 ≤ 5𝑥/5 < 10/5 –1 ≤ x < 2 Thus, x is a real number which is less than 2 and greater than or equal to –1 Hence, x ∈ [–1, 2) is the solution
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Chapter 6 Class 11 Linear Inequalities (Term 2)
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.