Question

find all parents of constructive 8 odd positive integer both of which are similar than 10 such that their sum is more than 11



PLEASE ANSWER MY QUESTION 50 POINTS ..........

Answer 1

(7,5) , (7,9) ,


explanation

Let the two consecutive odd positive integer be xx and x+2x+2.

Both number are smaller than 10 Therefore

x+2<10x+2<10

Adding −2−2 to both sides,

=>x<10−2=>x<10−2

=>x<8=>x<8

Also sum of the two integers is more than 11.

So, x+x+2>11x+x+2>11

=>2x+2>11=>2x+2>11

adding −2−2 to both sides,

=>2x>11−2=>2x>11−2

=>2x>9=>2x>9

Adding -2 to both sides,

=>2x>11−2=>2x>11−2

=>2x>9=>2x>9

Diving by 2 on both sides,

=>x>9/2=>x>9/2

=>x>4.5=>x>4.5

Step 2 :

Since x is an odd integer number greater than 4.54.5 and less than 8 (from 0) x can take values 5 and 7.

Thus the required pairs are (5,7)(5,7) and (7,9)(7,9)

Answer 2

Answer:

Step-by-step explanation:

x+2<10

Adding −2 to both sides,

=>x<10−2

=>x<8

Also sum of the two integers is more than 11.

So, x+x+2>11

=>2x+2>11

adding −2 to both sides,

=>2x>11−2

=>2x>9

Adding -2 to both sides,

=>2x>11−2

=>2x>9

Diving by 2 on both sides,

=>x>9/2

=>x>4.5