Question

The sum of three consecutive integers is 5 more than the smallest of the integers.find the integers

Answer 1

Let smallest no. is x
No. are x, x+1,x+2
Do x +x+1+x+2=x+5
3x+3=x+5
3x=x+5-3
3x=x+2
3x-x=2
2x=2
x=2/2
x=1

Let's verify=1+2+3=6
6-1=5
so it is correct answer
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Answer 2

Question:-

☛ The sum of three consecutive integers is 5 more than the smallest of the integers.Find the integers.

Given:-

☛ The sum of three consecutive integers is 5 more than the smallest integers.

To Find:-

☛ The three consecutive integers.

Unknown:-

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

Equation:-

☛ The sum of these integers will be x + x + 1 + x + 2 = 3x + 3.

☛ Since the sum of these integers is 5 more than the smallest integers,3x + 3 = 5 + x.

Solution:-

$\sf{3x + 3 = 5 + x} \\ \sf{3x + 3 - x = 5 + { { \cancel{x}}}} - { { \cancel{x}}} \\ \sf{2x + 3 = 5} \\ \sf{2x + { { \cancel{3}}}} - { { \cancel{3}}} = 5 - 3 \\ \sf{2x = 2} \\ \sf{ \frac{2x}{2} = { { \cancel \frac{2}{2}}}} \\ \sf{x = 1} \\ \sf{x + 1 = 1 + 1 = 2 \: and \: x + 2 = 1 + 2 = 3}$

Hence,the three consecutive integers are 1,2 and 3.

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