Question

The sum of three consecutive integers is 10 more than the twice of the smallest integer. Find the numbers.
pls answer quick need to submit homework urgently

Answer 1

• Numbers are 7, 8 and 9.

Step-by-step explanation:

To find :-

• Numbers.

Solution:-

Given that,

Sum of three consecutive integers is 10 more than the twice of the smallest integer.

Let, The three consecutive integers be x, x+1 and x+2.

According to question :

• The smallest integer is x. Twice of x = 2x.

➝ x + x + 1 + x + 2 = 10 + 2x

➝ 3x + 3 = 10 + 2x

➝ 3x = 10 + 2x - 3

➝ 3x - 2x = 10 - 3

➝ x = 7

Verification:-

➝ x + x + 1 + x + 2 = 10 + 2x

• Put x = 7

➝ 7 + 7 + 1 + 7 + 2 = 10 + 2×7

➝ 7 + 8 + 9 = 10 + 14

➝ 24 = 24

Hence, Verified!!

We have taken consecutive integers be x, x + 1 and x + 2. So,

x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Therefore,

Numbers are 7, 8 and 9.

Answer 2

Answer:

Given :-

• The sum of three of three consecutive integers is 10 more than the twice of the smallest integers.

To Find :-

• What is the number.

Solution :-

✦ Let, the first integers be x

✦ Second integers be x + 1

✦ And, the third integers be x + 2

According to the question,

⇒ x + (x + 1) + (x + 2) = 2x + 10

⇒ x + x + 1 + x + 2 = 2x + 10

⇒ x + x + x + 1 + 2 = 2x + 10

⇒ 3x + 3 = 2x + 10

⇒ 3x - 2x = 10 - 3

➠ x = 7

Hence, the required number,

❐ First number = x = 7

❐ Second number = x + 1 = 7 + 1 = 8

❐ Third number = x + 2 = 7 + 2 = 9

∴ The numbers are 7,8 and 9 .

_______________________