Twice the smallest of three consecutive odd integers is seven more than the largest, how do you find the integers?
Interpret the question and solve to find:
#11#, #13#, #15#
If the smallest of the three integers is
#2n = (n+4)+7 = n+11#
#n = 11#
So the three integers are:
The three consecutive odd integers are
We are given 3 consecutive odd integers.
Let the first odd integer be
Then the next odd integer will be
Now, We have our three integers,
Clearly the smallest integer is
Given that: twice the smallest = 7 more than the largest.
Our 3 consecutive odd integers are
Twice the smallest =
7 more than the largest =