Twice the smallest of three consecutive odd integers is seven more than the largest, how do you find the integers?

2 Answers
May 6, 2016

Answer:

Interpret the question and solve to find:

#11#, #13#, #15#

Explanation:

If the smallest of the three integers is #n# then the others are #n+2# and #n+4# and we find:

#2n = (n+4)+7 = n+11#

Subtract #n# from both ends to get:

#n = 11#

So the three integers are: #11#, #13# and #15#.

May 6, 2016

Answer:

The three consecutive odd integers are #11#, #13# and #15#.

Explanation:

We are given 3 consecutive odd integers.

Let the first odd integer be #x#.

Then the next odd integer will be #x+2#.

Since #x# is odd, #x+1# will be even, and we want 3 odd integers that are consecutive.

The #3^(rd)# integer will be #x+2+2=x+4#

Now, We have our three integers, #x#, #x+2# and #x+4#.

Clearly the smallest integer is #x# and the largest is #x+4#.

Given that: twice the smallest = 7 more than the largest.

#=>2x = 7 + (x + 4)#

#=>2x = x + 11#

#=>x = 11#

Checking

Our 3 consecutive odd integers are #11#, #13# and #15#.

Twice the smallest = #2xx11=22#

7 more than the largest = #7+15 = 22#