Question

Help for a medal! please!

Answer 1

Find three consecutive even numbers so that the sum of the first and third numbers is 22 less than three times the second number.

My brain can't handle this.

Answer 2

lol guess and check forget equations
you will probably get it on the first try

Answer 3

here, my guess is 6,8,10
i add and get
\[6+8+10=14\]

Answer 4

hmm idk. I think its a bit more complicated than that but ill take anything at this point! :P thanks!

Answer 5

oh i lied never mind
lets go ahead and solve it

Answer 6

*facepalm* okay...

Answer 7

lets call the first number \(n\) so the second one is \(n+2\) and the third one is \(n+4\)
so far so good?

Answer 8

sure

Answer 9

now we can translate
the sum of the first and third numbers
as
\[n+n+4\]

Answer 10

and we can translate
22 less than three times the second number
as
\[3(n+2)-22\]

Answer 11

we make "is" as \(=\) and the whole thing becomes
\[n+n+4=3(n+2)-22\] or
\[2n+4=3(n+2)-22\] and we can solve that one

Answer 12

you good from there?

Answer 13

umm I think so

Answer 14

\[2n+4=3n+6-22\]
\[2n+4=3n-16\]
\[2n+20=3n\]
\[20=n\]

Answer 15

is 20=n the final product?