The sum of three consecutive integers is 141. what is the smallest integer?

The Consecutive number are 46 , 47 and 48 having sum as 141

The smallest number is 46

Step-by-step explanation:

• A consecutive numbers means number are in a sequence and are one after the another !

e.g. n , (n+1). Or 1 , 2 are consecutive numbers

• A consecutive even numbers differ by 2 because all even numbers are divisible by 2 !

• If the number is divisible by x we must had taken number as n , ( n + x ) ,

( n + 2x )

• Let the numbers be ( n - 1 ) , n and

( n + 1 )

Given ,the sum of numbers is 141

n - 1 + n + n + 1 = 141

3n = 141

n = 141/3

n = 47

So, the numbers are

n- 1 = 47 - 1 = 46

n = 47

n + 1 = 48

And , the smallest number is 46

• Verification :-

46 + 48 + 47 = 141

Which satisfies the given condition !!

Hence , verified !!

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.