Find three consecutive odd integers such that their sum is fi ve more than four times the largest integer.

Question

I need help I can't remember how to do these.

Answer 1

Hi Joshua,

Let's use the variable, x , to represent the smallest of the three consecutive odd integers.

Consecutive odd integers are always 2 away from each other.

Since x represents the smallest integer here, the next consecutive odd integer would be written as x + 2 and the third integer as x + 4.

Next, the sum of these integers would be written as: x + x + 2 + x + 4

This is equivalent to 3x + 6

Since the problem states that this sum is 5 more than 4 times the largest integer, we can subtract 5 from the sum and set it equal to 4 times the largest integer:

4 times the largest integer would be written as 4( x + 4) = 4x + 16

Now we can create and solve the following equation to find the smallest integer:

3x + 6 - 5 = 4x + 16

3x + 1 = 4x + 16

3x + 1 - 3x = 4x + 16 - 3x

1 = 1x + 16

1 - 16 = 1x + 16 - 16

-15 = x

The smallest integer will be -15, The next bigger consecutive odd integer will be -13 followed by -11.

The integers will be -15, -13, -11

Their sum is -39. 4 times the largest is 4 • (-11) = -44

Note that -39 is 5 more than -44.