How do you find three consecutive odd integers such that he sum of the first and twice the second is 6 more than the third?

1 Answer
Jan 10, 2017

See process for finding the solution to this problem below:

Explanation:

First, let's define the three consecutive odd numbers.

The first, we can call #x#.

Because they are odd numbers we know they are every other number from #x# so we need to add #2#

The second and third number will be #x + 2# and #x + 4#

"twice the second" is the same as: #2(x + 2)#

"the sum of the first and twice the second is then: #x + 2(x + 2)#

This sum is "6 more than the third" or #(x + 4) + 6#

Putting this into an equation and solving for #x#:

#x + 2(x + 2) = (x + 4) + 6#

#x + 2x + 4 = x + 10#

#3x + 4 = x + 10#

#3x + 4 - color(red)(x) - color(blue)(4) = x + 10 - color(red)(x) - color(blue)(4)#

#3x - color(red)(x) + 4 - color(blue)(4) = x - color(red)(x) + 10 - color(blue)(4)#

#2x + 0 = 0 + 6#

#2x = 6#

#(2x)/color(red)(2) = 6/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 3#

#x = 3#

The first number is 3 and therefore the next two consecutive odd numbers are 5 and 7