When the smaller of two consecutive odd integers is added to four times the larger, the result is 30 more than 3 times the smaller. find the integers.?

2 Answers
LM
Jan 21, 2018

#11, 13#

Explanation:

denote the smaller integer as #n#, and the larger consecutive odd integer #n+2#.

smaller is added to four times the larger:

#n + 4(n+2)#

#30# more than #3# times the smaller:

#3n + 30#

this means that #n+ 4(n+2) = 3n + 30#

expand brackets:

#n + 4n + 8 = 3n + 30#

#5n + 8 = 3n + 30#

subtract #3n:#

#2n + 8 = 30#

subtract #8#:

#2n = 22#

divide by #2:#

#n = 11#

this means that the smaller integer is #11#

the larger integer is #n +2#, which is #13#.

Sean
Jan 21, 2018

#11# and #13#

Explanation:

.

Let's let #2n# be an even integer. The coefficient #2# ensures that the integer is always even regardless of whether #n# itself is odd or even.

Then, #2n+1# would always be an odd integer. Let's consider it to be the smaller one.

And #2n+3# would be the next consecutive integer (the larger one).

Four times the larger integer is:

#4(2n+3)#

Three times the smaller integer is:

#3(2n+1)#

Now, we can write an equation based on the wording of the problem:

#(2n+1) +4(2n+3)=3(2n+1)+30#

#2n+1+8n+12=6n+3+30#

#10n+13=6n+33#

#4n=20#

#n=5#

The smaller integer is:

#2n+1=2(5)+1=10+1=11#

The larger one is:

#11+2=13#

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