Find two consecutive even integers such that the sum of the larger and twice the smaller is 62?

Question

11747 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-15T04:16:46

#20 and 22#

Let the two consecutive even numbers be: #2n and 2(n+1)# for some #n in NN#

Since #n in NN -> 2(n+1) > 2n#

We are told that the sum of the larger plus twice the smaller equals 62.

Hence, #2(n+1) + 2xx 2n =62#

#2n+2+4n =62#

#6n = 60 -> n=10#

Replacing #n=10# into #2n and 2(n+1)#

#-># Smaller number #=20# and larger number #=22#

Check:

#20 and 22# are consecutive even numbers.

#22+2xx20 = 62#