Ex 6.1, 4 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Last updated at Feb. 15, 2020 by Teachoo
Transcript
Ex 6.1, 4 Solve 3x + 8 > 2, when x is an integer 3x + 8 > 2 3x > 2 โ 8 3x > โ6 ๐ฅ > (โ6)/3 ๐ฅ > โ2 Since x is an integer (โฆโฆ,โ3, โ2, โ1, 0, 1, 2, 3โฆ.) We need to find values of x which is greater than -2 i.e. x can be -1, 0, 1, 2, 3, 4,โฆโฆ.. = {โ1, 0, 1, 2, 3, 4,โฆโฆ} Integers: โฆ..,โ2, โ1, 0, 1, 2, 3,โฆ. Ex 6.1, 4 Solve 3x + 8 > 2, when (ii) x is a real number Now, x > โ2 Since x is a real number which is greater than โ2 Thus x โ (โ2, โ)
Ex 6.1
Ex 6.1, 2
Ex 6.1, 3
Ex 6.1, 4 Important You are here
Ex 6.1, 5
Ex 6.1, 6 Important
Ex 6.1, 7
Ex 6.1, 8 Important
Ex 6.1, 9
Ex 6.1, 10 Important
Ex 6.1, 11 Important
Ex 6.1, 12
Ex 6.1, 13
Ex 6.1, 14
Ex 6.1, 15
Ex 6.1, 16 Important
Ex 6.1, 17 Important
Ex 6.1, 18
Ex 6.1, 19
Ex 6.1, 20 Important
Ex 6.1, 21
Ex 6.1, 22 Important
Ex 6.1, 23 Important
Ex 6.1, 24
Ex 6.1, 25 Important
Ex 6.1, 26
Ex 6.1
About the Author