Find all pairs of consecutive even positive integers both of which are

1.	Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.
Answer» Let the pair of consecutive even positive integers be x and x + 2. So, it is given that both the integers are greater than 8 Therefore, x > 8 and x + 2 > 8 When, x + 2 > 8 Subtracting 2 from both the sides in above equation x + 2 – 2 > 8 – 2 x > 6 Since x > 8 and x > 6 Therefore, x > 8 It is also given that sum of both the integers is less than 25 Therefore, x + (x + 2) < 25 x + x + 2 < 25 2x + 2 < 25 Subtracting 2 from both the sides in above equation 2x + 2 – 2 < 25 – 2 2x < 23 Dividing both the sides by 2 in above equation \(\frac{2{\text{x}}}{2}< \frac{23}{2}\) x < 11.5 Since x > 8 and x < 11.5 So, the only possible value of x can be 10 Therefore, x + 2 = 10 + 2 = 12 Thus, the required possible pair is (10, 12).

Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.

Answer»

Let the pair of consecutive even positive integers be x and x + 2.

So, it is given that both the integers are greater than 8

Therefore,

x > 8 and x + 2 > 8

When,

x + 2 > 8

Subtracting 2 from both the sides in above equation

x + 2 – 2 > 8 – 2

x > 6

Since x > 8 and x > 6

Therefore,

x > 8

It is also given that sum of both the integers is less than 25

Therefore,

x + (x + 2) < 25

x + x + 2 < 25

2x + 2 < 25

Subtracting 2 from both the sides in above equation

2x + 2 – 2 < 25 – 2

2x < 23

Dividing both the sides by 2 in above equation

\(\frac{2{\text{x}}}{2}< \frac{23}{2}\)

x < 11.5

Since x > 8 and x < 11.5

So, the only possible value of x can be 10

Therefore, x + 2 = 10 + 2 = 12

Thus, the required possible pair is (10, 12).