Question

solve:
i have this question almost done, just not sure about the ending

Answer 1

trying to figure this out right now, @burhan101

Answer 2

thanks liza :) would you like me to rotate it ?

Answer 3

No, I figured it out. Is there a specific way that you were supposed to do this problem?

Answer 4

not but we just had to let it equal zero

Answer 5

is the way i have it right so far ?

Answer 6

OH, I'm sorry, I misunderstood, I thought you were solving for x.

Answer 7

no, the top is the x-intercept and the bottom gives us the restrictions

Answer 8

I have NO clue. Sorry, I'm out of my league!

Answer 9

it's okay

Answer 10

There's an easier way to do that.

Answer 11

I have to make the side equal zero tho

Answer 12

Why?

Answer 13

i don't know, that's the way my teacher wants it done :(

Answer 14

Well, it looks like what you did is fine for that method, just keep it =0 at the end.

Answer 15

\[\large \frac{-x+3}{4(3x-1)}=0\]

This looks ok, so go ahead and solve for x.

Answer 16

no, there's something wrong with it like that. one sec ill upload the picture with my teacher's comment on it

Answer 17

pfft, whatever, those are just stylistic differences. It won't make any difference to the solution.

Answer 18

Sure, you can take the -1 out and have (x - 3) on top if you want.

Answer 19

ohhh that's it ?

Answer 20

i got docked a whole mark for that -.- haha

Answer 21

Your teacher sounds way too picky. Here's how I would do it:
\[\large \frac{2x}{3x-1}=\frac{3}{4}\]
\[\large \rightarrow4(2x) = 3(3x-1)\]
\[\large \rightarrow 8x=9x-3 \rightarrow x=3, x \not=\frac{1}{3}\]

Answer 22

yes i do, i was just confused about my teachers comment, i thought there was more to it :$

Answer 23

that's so much simpleer @CliffSedge

Answer 24

Yeah, but if your teacher is making you do it a particular way and is grading on 'style' then I guess you have to do it his/her way... :-\

Answer 25

I suppose your teacher wanted you to preserve the denominator so you wouldn't lose track of your domain restrictions, but you always determine those from the original equation anyway.

Answer 26

yeah i think that was her reasoning behind it i guess

Answer 27

@nincompoop burhan's teacher is picky about style and marked off points for order of multiplication of all things!

Answer 28

you can also consider these as an equation of a line or curve
\[\frac{ -x+3 }{ 4(3x-1) }=0\]
here you have a vertical line of x=1/3 and also a curve line that has a root of x=3

Answer 29

@mark_o. yes, that would be the first steps to graphing it, but I don't think that was a requirement of this problem.

@nincompoop check the pics burhan uploaded.

Answer 30

thank you so much for your help @CliffSedge, i am just making corrections to the mistake i have made on my test

Answer 31

I don't think you made a mistake; I think your teacher is just picky about how to do things.

Answer 32

Exactly!

Answer 33

either way is correct...lol

Answer 34

Again, pfft, I say, pfft.

Answer 35

that's true :P

Answer 36

LOL