solve 4/x-4=x/x-4-4/3 for x and determine if the solution is extraneous or not.
@zepdrix ^-^
Use LaTeX to post the equation.
ok just gimme a sec
Stop being so sloppy >:( At least use brackets lol
\[\frac{ 4 }{ x-4 } = \frac{ x }{ x-4 } - \frac{ 4 }{ 3 }\]
lol I don't wanna make you wait though ;-; so i just end up doing the equation maker after
You want to multiply both sides of the equation by the `Least Common Multiple`. Don't worry about common denominator, just multiply all the denominators out of there.
4/(x - 4) = x/(x -4) - 4/3 4/(x - 4) - x/(x - 4) = - 4/3 ( 4 - x)/(x - 4) = - 4/3 - (x - 4)/(x - 4) = - 4/3 no solution x = 4, extraneous
ok....well I'll do the work but thanks chealyn lol
btw zep, can I ask oneee more question?
I'll use brackets this time babe no worries lol
Ok
What is the simplified form of \[\frac{ 3 }{ 4x+3 } + \frac{ 21 }{ 8x^2 + 26x + 15 }\]
uhh I dunno, it looks right to me
I guess you gotta factor some stuff, ya?
8*15 = 120 ish
what
Oh, we have a "hint", Notice the other denominator is 4x+3, so this quadratic in the other denominator PROBABLY factors into (4x+3) times something else.
ok...can i show you the options? I think it's B or C but lemme post it
dunno tbh this one is hard
\[\large\rm 8x^2+26x+15\]You'll have to factor this one by grouping. 8*15 = 120. We need factors of 120 that `add` to 26.
\(\rm 2*60 = 120\) \(\rm 2+60 \ne26\) Hmm those didn't work.
ugh ok lemme think
3 * 40 no
4 + 30
* 4 * 30
so -4, +30
No, it's a positive 120, and positive 26, so they both have to be positive
4 and 30 are getting closer though, right? 34 is close to 26, so we're "close to" the correct factors.
The numbers need to be just a tad closer together.
um
i mean cant you do 30-4
oh wait
its all positive
nvm lol ok so it's 6 and 20
Good good good. So that's how we'll break down our 26, into 20 and 6.\[\large\rm 8x^2+\color{orangered}{26x}+15\]\[\large\rm 8x^2+\color{orangered}{20x+6x}+15\]And now factor by grouping,\[\large\rm (8x^2+20x)+(6x+15)\]
Remember this process? Lil bit?
yess
but idk what to do now
What can you factor out of the first "group"? They each have an x in them, what else they sharing?
an even number? so it'd be 4x^2 + 5x
I'm not just what you just did -_- You can take an x out of each,\[\large\rm (8x^2+20x)=x(8x^2+20)\]And certainly a 2 out of each (which maybe was what you were trying to do...), but I think you can take a larger value than a 2 out.
oh my god I just wanna hit myself in the face rn I'm so dumb. It's 11:20 pm rn tho so can i just spit out a guess and you tell me if i'm right?
woops, I took the x out incorrectly :3 hope you noticed that
i did lol
\[\large\rm (8x^2+20x)=4x(2x+5)\]
is it D?
cos the (2x+5)
\[\large\rm (8x^2+20x)+(6x+15)\]\[\large\rm 4x(2x+5)+(6x+15)\]Pulling a 3 out of the other one,\[\large\rm 4x(2x+5)+3(2x+5)\]Factoring further,\[\large\rm (4x+3)(2x+5)\]
Woman I don't know -_- Don't be lazy lol
it is D!! hahahhaa
aww Zep I'm sorry ;-;
\[\large\rm \frac{ 3 }{ 4x+3 } + \frac{ 21 }{ 8x^2 + 26x + 15 }\]So our expression becomes,\[\large\rm \frac{ 3 }{ 4x+3 } + \frac{ 21 }{(4x+3)(2x+5)}\]
So for common denominator, you need a (2x+5) in the first fraction.
yeayea i just put it in as the answer. anyway thanks so much man!! you're my favorite on here ^-^ have a nice night bb okie? and enjoy the winter weather omg idk how it is in daytona but it's greaaat where I'm at
night~
...le sigh -_-
you .. too -_-
Oh you in FL too? :d It sucks here right now. Down in the 50's today D: One day of cold and I'm already hating it haha.
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