find three consecutive odd integers such that twice the sum of the first and third integers is 21 more than the second integer. Define a variable, set up an equation and solve it.

X is my variable for my first odd integer. Since it must be consecutive the second variable will be x+2 and the third will be x+4

1st int = x2nd int = x+23rd int = x+4

And 2*( x+x+4)=21+x+22 * (1st int + 3rd int) = 21+ second int

simplify

2x+2x+8=21+x+2

4x-x=21+2-8

3x=15

x=5

1st int =5=x

2nd int =x+2= 5+2=7

3rd int = x+4=5+4=9

The answer is 5,7,9 <------

verifying: 2*(5+9)=21+7

2*(14)=28

28=28

hope this helps

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