Question

PLS HELP ME IM DYING

find three consecutive even integers such that the sum of the smallest and 3 times the second is 26 more than twice the third!

The three consecutive integers are:

Answer 1

hmmm say if you pick any integer whatsoever, and multiply it by some even value, the product will be an even integer, for example, say "3", 2(3) = 6 even, or say "8", 2(8) = 16 even again, so let's say the 1st even integer we'll use is "2a", to get the consecutive ones, we can simply either add or subtract 2 from that.

`\begin{array}{rllll} 2a&=&\textit{first integer}\\\\ 2a+2&=&\textit{second consecutive integer}\\\\ 2a+4&=&\textit{third consecutive integer} \end{array}`

`\stackrel{smallest}{2a}~~ + ~~\stackrel{\textit{three times the 2nd}}{3(2a+2)}~~ = ~~\stackrel{\textit{twice the 3rd}}{2(2a+4)}+\stackrel{\textit{and this more}}{26} \\\\\\ 2a+6a+6~~ = ~~4a+8+26\implies 8a+6~~ = ~~4a+34 \\\\\\ 4a+6=34\implies 4a=28\implies a=\cfrac{28}{4}\implies a=7 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \underset{first}{\stackrel{2(7)}{14}}\hspace{5em}\underset{second}{\stackrel{2(7)~~ + ~~2}{16}}\hspace{5em}\underset{third}{\stackrel{2(7)+4}{18}}~\hfill`

Answer 2

14, 16, 18

sum of the smallest (14) and 3 times the second (16)

14 + 3 x 16 = 62

is 26 more than twice the third (18)

2 x 18 = 36

36 + 26 = 62