Question

I need help with Algebra 2. :c Any takers? Questions in comments.

Answer 1

this is a pain, but we can do it

Answer 2

which one you want to start with?

Answer 3

actually only the first one is a pain, the last two you do with your eyeballs

Answer 4

Hopefully! I've been stuck on them all day. I want to start with the one on top. The solving one.

Answer 5

ok that is the only hard one

Answer 6

\[\frac{x-5}{3x}<3\] you CANNOT multiply both sides by \(3x\) because it could be positive or negative, so you have to subtract \(3\) from both sides get
\[\frac{x-5}{3x}-3<0\]

Answer 7

then you actually have to do the subtraction
\[\frac{x-5-9x^2}{3x}<0\]

Answer 8

there has to be an easier way let me think for a second

Answer 9

Okay! Thank you so much!

Answer 10

i mean this will work, no problem, just seeing if i can find a simpler method

Answer 11

I have one more problem exactly like that one, so a simpler method would be much appreciated. c:

Answer 12

ok lets do it this way
work in cases
if \(x>0\) then we can multiply both sides by \(3x\) and get
\[\frac{x-5}{3x}<3\\
3x^2-x+5>0\]

Answer 13

\[y=3x^2-x+5\] is a parabola that opens up
the discriminant it negative, meaning it has not real zeros, meaning it is always positive
so one solution is \(x>0\)

Answer 14

now we do it if \(x<0\) so we switch the inequality when multiplying both sides by \(3x\) and get
\[x-5>3x^2\] or \[-3x^2+x+5<0\] damn i must be doing something wrong here

Answer 15

give me a second i have to write it on paper

Answer 16

ok are you still there

Answer 17

i feel like an idiot

Answer 18

I am

Answer 19

good lets go back and do it the correct way

Answer 20

\[\frac{x-5}{3x}<3\\
\frac{x-5}{3x}-3<0\] now lets subtract correctly
\[\frac{x-5-9x}{3x}<0\\
\frac{-8x-5}{3x}<0\]

Answer 21

i subtracted incorrectly like an idiot
now it is easy
the denominator changes sign at 0, the numerator changes sign at \(\frac{5}{8}\)

Answer 22

actually at \(-\frac{5}{8}\) since if you solve
\[-8x-5=0\] you get
\[-8x=5\\
x=-\frac{5}{8}\]

Answer 23

Oh I see now!

Answer 24

so you have three intervals
\[(-\infty, -\frac{5}{8}), (-\frac{5}{8}, 0), (0,\infty)\] and you just need to check the sign on one of them

Answer 25

i believe it will be \(-,+,-\) and you want \(-\) because we were solving \[
\frac{-8x-5}{3x}<0\] less then zeros being a synonym for negative

Answer 26

making your answer
\[(-\infty, -\frac{5}{8})\cup (0,\infty)\] sorry i got off to a false start with my bad algebra

Answer 27

we can do another if you like just to make it clear, and clear that it is not that hard

Answer 28

It's okay. c: Thanks for helping me

Answer 29

or we can look at your next two, which are eyeball problems

Answer 30

for B
what is the y coordinate on the curve at \(x=6\)?

Answer 31

I finished the other two. I had a brain fart and realized how to do them cx

Answer 32

oh ok i thought they were easy

Answer 33

There's another problem like the one you just did, if you're willing to help out again

Answer 34

want to solve another inequality quickly? make sure it is obvious (feel guilty about messing up such an easy question)

Answer 35

yeah sure lets do it

Answer 36

\[\frac{1}{b}=\frac{6}{5b}+1\]?

Answer 37

this is an equality, right, not an inequality
so this time you can multiply both sides by \(5b\) and start with
\[5=6+5b\]
that should be easy to solve

Answer 38

yes

Answer 39

equalities are much much easier (always) than inequalities
you good from there?
two steps
subtract 6,
divide by 5

Answer 40

Okay. I guess they seem harder to me than they should be cx

Answer 41

three were easy enough, the inequality required more careful algebra
that it?

Answer 42

yep. That's it. c: