Question

Let's write down three consecutive integers such that 5 less than the sum of the
numbers is equal to 11 more than twice the second number. Let's find the three
consecutive integers. ​

Answer 1

Answer:

Let the three consecutive numbers be x, (x+1), (x+2).

Now according to the given criteria,

5 less than the sum of the number = x+(x+1)+(x+2)-5…..(i)

11 more than twice the second number = 2(x+1)+11…….(ii)

So from question equation (i) and equation (ii) are equal,

Hence

x+(x+1)+(x+2)-5 = 2(x+1)+11

⇒ 3x+3-5=2x+2+11

⇒ 3x-2=2x+13

⇒ 3x-2x=13+2

⇒ x=15

⇒ x+1=15+1=16

⇒ x+2=15+2=17

Hence the three consecutive numbers are

15, 16 and 17

Step-by-step explanation:

Pls mark me brainliest