View Full Version : Algebraic Problem
xxpsychoxx
Mar27-04, 10:12 PM
This question kinda stumped me. Can any1 post the answer with the working and all? Thanks :biggrin:
Show that
\frac{ 3 \left( \frac{x+5}{x-1} \right) + 4 }{ 4 \left( \frac{x+5}{x-1} \right) + 1 } = (3x+11)/{5x+19)
Sorry, we don't do your homework for you.
Anyways, you seem to be missing parentheses, did you mean:
\frac{ 3 \left( \frac{x+5}{x-1} \right) + 4 }
{ 4 \left( \frac{x+5}{x-1} \right) + 1 }
?
Anyways, what have you tried to do to solve this problem?
xxpsychoxx
Mar27-04, 11:06 PM
Sorry, we don't do your homework for you.
Anyways, you seem to be missing parentheses, did you mean:
\frac{ 3 \left( \frac{x+5}{x-1} \right) + 4 }
{ 4 \left( \frac{x+5}{x-1} \right) + 1 }
?
Anyways, what have you tried to do to solve this problem?
Yes, thats what i meant, but i was unsure on how to write it in that format. Well first, i tried to cross multiply but thats the problem..I'm not sure how to..
Well, in general, cross multiplication says that the equation
p/q = r/s
is equivalent to
ps = qr (with q and s inequal to 0)
Are you having trouble seeing how to make this substitution, or is it the next steps?
xxpsychoxx
Mar27-04, 11:22 PM
yep, it's the substitution thats giving me the problem so far
4((x+5/(x-1))+1 * 3x+11 only this one though
You have to put [ tex ] and [ /tex ] tags (no spaces) around the LaTeX code.
You're missing the ) after x + 5, but I think that's just a typo.
The big thing that you might be doing wrong is that you didn't put parentheses around each of the terms there. What you want is
( 4 ( \frac{x+5}{x-1} ) + 1 ) (3x + 11)
xxpsychoxx
Mar27-04, 11:39 PM
(4 ( \frac{x+5}{x-1} + 1 ) (3x+11)
hmm ok i think i got the hang of it :cool:
so can you tell me exactly how do i go about multiplying these two terms?
You're missing a parenthesis again!
There are at least two ways to progress from here.
One way is to look at the equation as a whole imagine the complicated thing is replaced by a simple thing; do you know how to expand (4z+1)(3x+11)?
The other way is to look at little pieces. Do you know any way to combine 4 (x+5)/(x-1) + 1 into one term?
xxpsychoxx
Mar27-04, 11:54 PM
hmmm i expanded (4x+1)(3x+11) and got 12x2+47x+11
I'm unsure about the other method
Try the same procedure, but on (4z+1)(3x+11) instead of (4x+1)(3x+11).
(We are justified in making a new letter to represent the fraction (x+5)/(x-1), but it has to be a new letter; replacing the fraction with x won't work)
xxpsychoxx
Mar28-04, 12:02 AM
Oh ok, i thought that was a typo :rolleyes:
Ok i got 12xz+44z+3x+11
That looks right. Now, since z was a substitute for the fraction (x+5)/(x-1), if you substitute the fraction back in for z, you will have successfully multiplied the two terms you had trouble with!
xxpsychoxx
Mar28-04, 12:20 AM
Ohhhhh ok i see :biggrin: Lol i didn't think of it that way. However I am still in a bit of a jam :confused: So i'm left with:
15x-5({x+5}/{x+1}) = 12x({x+5}/{x-1}) + 44({x+5}{x-1}) + 3x + 1
Or something like that :frown:
Well, there are (at least) again two approaches. :)
(a) Combine each side into a single fraction
(b) Clear the denominators (by multiplying both sides by the least common denominator)
And I'm off to bed.
xxpsychoxx
Mar28-04, 10:59 AM
hmmmm, my previous equation was incorrect. Here is the correct one:
57(x+5)/(x-1) + 15x(x+5)/(x-1) +20x + 76 = 12x(x+5)/(x-1) + 44(x+5)/(x-1) + 3x + 11
Then i subtract then get:
13(\frac{x+5}{x-1}) + 3x(\frac{x+5}{x-1}) = -17x -65
=\frac{13x(x+5) +3x(x+5)}{x+1}
Any errors? :redface:
xxpsychoxx
Mar28-04, 02:15 PM
\frac{13(x+5) +3x(x+5)}{x-1}
Oops sorry thats what i meant! OHHHHHHHHH YAY!! Thanks a lot ,i'm finally seeing the answer :biggrin: :biggrin: :biggrin: I got x is either equal to 0 or -3.8 :smile:
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