Question

Find three consecutive even integers such that three times the sum of the first and third is 24 greater than 4 times the second.

Answer 1

You want to find \(n, (n+1), (n+2)\) such that \(3(n+(n+2)) = 24+ 4(n+1)\). Can you solve this for n?

Answer 2

no... lol I'm not that good at math

Answer 3

try multiplying a bit

Answer 4

and adding like terms

Answer 5

for example n+(n+2) is the same as n+n+2
which is the same as?

Answer 6

2n + 2?

Answer 7

yes not do the multiplying part
there is lots of places where you can use distributive property
that is if you have a(b+c)
then this is equal to a*b+a*c

Answer 8

\[3(2n+2)=24+4(n+1)\]

Answer 9

now*

Answer 10

now what? lol

Answer 11

so you have done the distributing part? and you are asking for what step next?
if so please show what you have...

Answer 12

6n + 6 = 24 + 4n + 4

Answer 13

great
now collect like terms because you have some on the right hand side

Answer 14

6n + 6 = 28 + 4n

Answer 15

right now get your n terms on one side
and your terms without n on the opposing side

Answer 16

do this by adding/subtracting terms on both sides

Answer 17

2n = 22
?

Answer 18

6n-4n=28-6
2n=22
is right

Answer 19

n = 11 right?

Answer 20

yes

Answer 21

and oops
I just realized the problem said even

Answer 22

the word even is very important here

Answer 23

\[n,n+2,n+4\]

Answer 24

Oops, my bad.

Answer 25

\[3(n+[n+4])=24+4(n+2)\]
solve for n all over again
lol

Answer 26

its fine
those tiny words get looked over sometimes
I should have checked

Answer 27

would it be
6n + 4 = 24 + 4n + 8?

Answer 28

no
3(2n+4)=6n+12 not 6n+4

Answer 29

oh ok

Answer 30

6n + 12 = 24 + 4n + 8

Answer 31

yes

Answer 32

6n + 12 = 32 + 4n

Answer 33

great too :)

Answer 34

you are almost there

Answer 35

2n = 20

Answer 36

n = 10?

Answer 37

right
n=10
n+2=10+2=12
n+4=10+4=14
so the three numbers in question are 10,12,14

Answer 38

Thank you so much!

Answer 39

I don't think @AlixxKK would benefit from a Mathematica script; it isn't really suitable for this level.