Question

The sum of three consecutive odd integers is 41 less than four times the largest. Find the numbers.

Answer 1

Let n = 1st odd integer
   n + 2 = 2nd odd integer
   n + 4 = 3rd odd integer
SUM:
n + n + 2 + n + 4 = 4(n + 4) - 41
3n + 6 = 4n + 16 - 41
3n + 6 = 4n - 25
31 = n
ANSWER: 31, 33, 35

Answer 2

Consecutive numbers are numbers that follow one another.

The numbers are 27, 29 and 31

Let the numbers be x, y, z; where z is the largest.

Their sum is represented as:

$x + y + z = 4z - 41$

The numbers are then represented as:

$y = x + 2$

$z= x + 4$

Substitute these values in $x + y + z = 4z - 41$

So, we have:

$x + x + 2 + x + 4 = 4(x + 4) - 41$

Open brackets

$x + x + 2 + x + 4 = 4x + 16 - 41$

Collect like terms

$x + x + x - 4x = 16 - 41 - 2$

$-x = -27$

Rewrite as:

$x = 27$

Recall that:

$y = x + 2$

$z= x + 4$

$y = 27 + 2 = 29$

$z = 27 + 4 = 31$

Hence, the numbers are 27, 29 and 31

Read more about consecutive numbers at:

https://brainly.com/question/2493629