Question

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

Answer 1

Step-by-step explanation:

let the integers be x, x+1,x+2

x+x+1+x+2=2x+10

3x+3=2x+10

3x-2x=10-3

x=7

Answer 2

Answer:

$\large{ \boxed{ \underline{ \rm{7,8,9}}}}$

• The integers are 7,8 and 9

Step-by-step explanation:

Given :

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

To Find :

• The integers.

Solution :

Let's assume the integers as :-

• x
• x + 1
• x + 2

So, firstly let's calculate the value of x,

$\implies \rm \: x +( x + 1) + (x + 2) = 10 + 2x \\ \\ \\ \implies \rm 3x + 3 =10 + 2x \\ \\ \\ \implies { \boxed{\bf \: x = 7}}$

Hence, value of x is 7.

Now, let's calculate the remaining numbers :

$\tiny \dag \: { \underline{ \bf{First \: case}}}$

$\implies \rm \: x + 1 \\ \\ \\ \implies \rm \: 7 + 1 \\ \\ \\ \implies \bf \: 8$

Therefore, the second number is 8.

$\tiny \dag \: { \underline{ \bf{Second \: case}}}$

$\implies \rm \: x + 2 \\ \\ \\ \implies \rm \: 7 + 2 \\ \\ \\ \implies \bf \: 9$

Therefore, the third number is 9.

Henceforth,

• The integers are 7,8 & 9