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?

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Answer 1 · 2017-01-10T15:19:15

See process for finding the solution to this problem below:

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