How do you find three consecutive even integers such that three times the largest is 34 more than the sum of the two smaller integers?

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Answer 1 · 2017-07-04T02:52:31

See a solution process below:

Let's call the smallest of the three consecutive integers: #n#

Because these are consecutive even integers it means we can write the other two integers as:

#n + 2#

#n + 4#

Now, we can write "three times the largest" as:

#3(n + 4)#

And if this is equal to "34 more than the sum of the two smaller integers" we can write this as the equation:

#3(n + 4) = 34 + n + n + 2#

We can now solve for #n#:

#3(n + 4) = 34 + 1n + 1n + 2#

#3(n + 4) = 1n + 1n + 2 + 34#

#3(n + 4) = (1 + 1)n + 36#

#color(red)(3)(n + 4) = 2n + 36#

#(color(red)(3) * n) + (color(red)(3) * 4) = 2n + 36#

#3n + 12 = 2n + 36#

#-color(blue)(2n) + 3n + 12 - color(red)(12) = -color(blue)(2n) + 2n + 36 - color(red)(12)#

#(-color(blue)(2) + 3)n + 0 = 0 + 24#

#1n = 24#

#n = 24#

Therefore the three consecutive even integers are:

#n = 24#

#n + 2 = 26#

#n + 4 = 28#