Question

Question 24 Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Class X1 - Maths -Linear Inequalities Page 122

Answer 1

Let 2x and 2x +2 are two consecutive even positive integers .
Then a/C to question,
2x > 5 => x> 5/2

and 2x + 2 > 5
=> 2x > 5-2 = 3
=> 2x > 3
=> x> 3/2

and , sum of 2x and 2x +2 < 23
2x + 2x +2 < 23
=> 4x < 23 -2
=> 4x < 21
=> x < 21/4
now plotting these all value on number line , ( see attachment )
from attachment , its clear that
5/2 < x < 21/4
hence, possible value of x = 3, 4, 5
so, when x = 3 then, (2×3,2×3+2) = (6,8)
when x = 4 then , (2×4,2×4+2) = (8,10)
when x= 4 then , (2×5,2×5+2) = (10,12)
hence, required pair (6,8) , (8,10) and (10,12)

Answer 2

Answer:

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abhi178

Let 2x and 2x +2 are two consecutive even positive integers .

Then a/C to question,

2x > 5 => x> 5/2

and 2x + 2 > 5

=> 2x > 5-2 = 3

=> 2x > 3

=> x> 3/2

and , sum of 2x and 2x +2 < 23

2x + 2x +2 < 23

=> 4x < 23 -2

=> 4x < 21

=> x < 21/4

now plotting these all value on number line , ( see attachment )

from attachment , its clear that

5/2 < x < 21/4

hence, possible value of x = 3, 4, 5

so, when x = 3 then, (2×3,2×3+2) = (6,8)

when x = 4 then , (2×4,2×4+2) = (8,10)

when x= 4 then , (2×5,2×5+2) = (10,12)

hence, required pair (6,8) , (8,10) and (10,12)

Step-by-step explanation: