Question

3n+8<2(n-4)-2(1-n)

Answer 1

The first step would be to distribute the 2 and -2 through. From there, you would have to simplify the right side and then just solve for n on one side.

Answer 2

3n+8(is less than or equal to< with the line under it)2n-4-2+2n

is that correct?

Answer 3

Almost! Your first distribution is 2 into (n-4). You got 2n-4?

Answer 4

yeah i got that

Answer 5

If you're distributing, you'd also multiply the 4 by 2.

Answer 6

oh dang it my bad forgot to do that

Answer 7

2n-8 got it!

Answer 8

Yes. :)

So, you'd have something like...
\[3n-8 \le 2n-8-2+2n \]

Then, you combine like terms and it should be like any 'solve for n' problem.

Answer 9

er, it would be 3n+8, but yeah...

Answer 10

are the like terms -8-2=-10?

Answer 11

That is one set of the like terms. We also have 2 terms with 'n' on them, so we can also add those.

Answer 12

3n+8<2n-10+2n

Answer 13

2n + 2n = 4n.

Answer 14

3n+8<2n-10+2n
+2n +2n
3n+8<-10+4n , on the right track so fare

Answer 15

Yes. Now, you can work your way to solving for n

Answer 16

3n+8<-10+4n do i subtract 4n or 3n?

Answer 17

You are able to subtract either one. ( I would prefer subtracting 3n from 4n so you get a positive n, so you wont need to eventually switch the signs.)

Answer 18

Luis, you missed the "2" for "2(n-4)" :p

Answer 19

haha alright we are getting there, 3n+8<-10+n , thats what i got so fare!

Answer 20

Oh, you'd also subtract the 3n from the left side, so you keep them equal (or as equal as the inequalities can be)

Answer 21

8<-10+n

Answer 22

Basically, if you ever add or subtract something new on one side, you will always add or subtract the same thing on the other side.

Answer 23

and add 10?

Answer 24

Yes, add 10 to both sides

Answer 25

alright got is n is greater than or equal to 18!

Answer 26

it*

Answer 27

Yep. That's what I got as well. :)

\[
\begin{split}
3n+8 &\le 2(n-4)-2(1-n)\\
3n+8 &\le 2n-8 -2+2n\\
3n+8 &\le 4n-10\\
8 &\le n-10\\
18 &\le n\\
Or, n &\ge 18\\
\end{split}
\]

Answer 28

yup me to, thanks again!

Answer 29

your a math genus! haha

Answer 30

I hope I am helping you understand it as well. It's really nice to know this stuff for later math classes down the road. :D

Answer 31

Yeah i like how you are explaining it to me, hey is it alright if you help me out on a few more problems?

Answer 32

I could probably help for one more problem. I actually have some work I have to get back to myself. :p

Answer 33

\[k-17\le-(17-k)\]

Answer 34

Well, we first would distribute that -1.

\[k - 17 \le -17 + k \]

\[k-17 \le k - 17 \]

From what I can see here, both sides are exactly the same. So, for ANY k we use, we would get a true inequality.

So, we would have infinite solutions to this inequality. k can be any value for this to be true!