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# Simplifying this expression: Where have I gone wrong?

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# Simplifying this expression: Where have I gone wrong?

07-02-08, 03:51 PM
#1
phantomcow2
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Simplifying this expression: Where have I gone wrong?

What I did:

First, find common denominator of top fraction: a*b. Bottom fraction has the same common denominator. So multiplying out, I get....

((b-a^2)-(b^2+a))/ab

And for hte bottom fraction: ((b^2-a)+(b+a^2))/ab

But since these two fractions must be divided, I take the reciprocal. Making.......

((b-a^2)-(b^2+a))/ab * ab/((b^2-a)+(b+a^2))

At this point, don't the ab terms cancel? Leaving me with
((b-a^2)-(b^2+a))/((b^2-a)+(b+a^2))
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07-02-08, 04:15 PM
#2
jschen
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PC2, does (b-a^2)/ab equal (b-a)/b ?

You see what you did wrong yet?

If not, please show your work "multiplying out".
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Last edited by jschen; 07-02-08 at 04:19 PM.
07-02-08, 04:17 PM
#3
jschen
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There is also another point I'd like to discuss with you about your choice of methods for simplification, but after we resolve the more serious issue. What you did in this case (edit: what I hope to discuss later, not what I mentioned above) isn't wrong, but there is a more natural and straightforward method that one should train oneself to easily see at first glance.

But anyway, let's focus on what's wrong before focusing on efficiency/elegance.
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Last edited by jschen; 07-02-08 at 04:21 PM.
07-02-08, 04:20 PM
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SwimBike
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wow...I dont remember any of this stuff anymore.
07-02-08, 04:20 PM
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It's math. That is where you went wrong.
07-02-08, 04:24 PM
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This is a classic.

You only did half of the multiplying you were supposed to do. You were so concerned about getting rid of the a*b in the denominators, you forgot to keep things "balanced", and do the same thing to the numerators.

Hint:

a/b - c/d =

(d/d)*(a/b) - (b/b)*(c/d) =

(a*d - bc)/bd

Where a and c are arithmetic equations in your problem.
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07-02-08, 04:24 PM
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jschen
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By the way, a and b are simply numbers that can vary (ie variables). If you still don't follow my first response, or if you figure that out but get stuck again elsewhere, try substituting, say, a = 1 and b = 2 to see at which step things go downhill.
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07-02-08, 05:19 PM
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apclassic9
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does this have something to do with your car's overheating problem?
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07-02-08, 05:37 PM
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The answer is -1.

If I didn't make a mistake, all the a's and b's cancel out.
07-02-08, 05:44 PM
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You forgot to carry the three.
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07-02-08, 05:51 PM
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deraltekluge
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Starting with

Multiply by ab/ab

The numerator becomes a(b-a) - b(b+a)

The denominator becomes b(b-a) + a(b+a)

Expand those, and you get (ab - a² -b² - ab)/(b² - ab + ab + a²)

Or, -(a² + b²)/(a² + b²) = -1
07-02-08, 05:58 PM
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deraltekluge, you are correct. Phantomcow2, I think what jschen will probably tell you is that you're overthinking the problem. In this case, don't think about finding common denominators, it will just complicate things. Multiply things out and (I hope) you'll see the answer. It's pretty straightforward, you're just overthinking it.
07-02-08, 06:00 PM
#13
huytheskigod
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Originally Posted by deraltekluge
Starting with

Multiply by ab/ab

The numerator becomes a(b-a) - b(b+a)

The denominator becomes b(b-a) + a(b+a)

Expand those, and you get (ab - a² -b² - ab)/(b² - ab + ab + a²)

Or, -(a² + b²)/(a² + b²) = -1
Don't just give it to him...how is he going to learn if you do that....
07-02-08, 06:05 PM
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deraltekluge
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Originally Posted by phantomcow2
So multiplying out, I get....

((b-a^2)-(b^2+a))/ab

And for hte bottom fraction: ((b^2-a)+(b+a^2))/ab

You multiplied a*(b - a) and got (b - a²) instead of (ab - a²)

And you did that same sort of thing all four times.
07-02-08, 06:18 PM
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deraltekluge
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If you were averse to squaring things, you could recognize that

(b - a)/b = 1 - a/b

Do that to all four terms, and you get

(1 -a/b - b/a -1)/(b/a -1 + 1 + a/b) = (-a/b - b/a)/(b/a + a/b) = -(a/b +b/a)/(a/b + b/a) = -1
07-02-08, 06:34 PM
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jschen
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Originally Posted by huytheskigod
deraltekluge, you are correct. Phantomcow2, I think what jschen will probably tell you is that you're overthinking the problem. In this case, don't think about finding common denominators, it will just complicate things. Multiply things out and (I hope) you'll see the answer. It's pretty straightforward, you're just overthinking it.
Yes, deraltekluge's work is correct. I don't think PC2 overthought it. He simply made a mistake in multiplication that he shouldn't make if he weren't rusty with his algebra. That said, once he recognized both common denominators as ab, rather than taking reciprocals, he should have simply multiplied numerator and denominator by ab. Mathematically the same thing as what he did, but a bit prettier and quite a bit shorter to write everything out.
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07-02-08, 06:36 PM
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jschen
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Originally Posted by deraltekluge
If you were averse to squaring things, you could recognize that

(b - a)/b = 1 - a/b

Do that to all four terms, and you get

(1 -a/b - b/a -1)/(b/a -1 + 1 + a/b) = (-a/b - b/a)/(b/a + a/b) = -(a/b +b/a)/(a/b + b/a) = -1
Cool! I think that's a rather elegant solution. Not obvious at first glance that it wouldn't create more of a mess, but it doesn't; indeed, it keeps things very simple.

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07-02-08, 08:01 PM
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Fricken over-achievers.

07-03-08, 02:18 PM
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07-03-08, 02:29 PM
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07-03-08, 02:35 PM
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Originally Posted by phantomcow2
Simplifying this expression: Where have I gone wrong?
please, for the love of god, find yourself a legally aged woman, who's willing to have intimate relations with you. ASAP.
07-03-08, 03:02 PM
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crash66
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Originally Posted by botto
please, for the love of god, find yourself a legally aged woman, who's willing to have intimate relations with you. ASAP.

That cutie actress from the old Wonder Years show is a math brainiac.

Any LA based BFers know her? Can you do phantom a solid and hook a brother up???
07-04-08, 08:35 AM
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Originally Posted by jschen
indeed, it keeps things very simple.
HUH?
07-05-08, 03:33 PM
#24
phantomcow2
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I should have reported back earlier. Thanks, this problem might make sense now. But there is another congregation of impending frustration, which I must solve:
https://web-homework.sr.unh.edu/webwo...d7942630b1.png

So the best I can think to do is square everything, denominator and numerator. THat leaves me with......
(6-2x)+(25/4x)1/(6-2x)^-4 / (6-2x)^2

6-2x^2 "FOIL's" out to 4x^2-24x+36, but I don't see how that would help me anyways.

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07-05-08, 10:00 PM
#25
FlowerBlossom
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Originally Posted by phantomcow2
I should have reported back earlier. Thanks, this problem might make sense now. But there is another congregation of impending frustration, which I must solve:
https://web-homework.sr.unh.edu/webwo...d7942630b1.png

So the best I can think to do is square everything, denominator and numerator. THat leaves me with......
(6-2x)+(25/4x)1/(6-2x)^-4 / (6-2x)^2

6-2x^2 "FOIL's" out to 4x^2-24x+36, but I don't see how that would help me anyways.

I haven't checked your work, so I don't know if you have the correct answer.

However, typically, you are doing these kinds of problems as a set up-for finding the values that make the equation = 0. Note the plural, "values". There can be more than one value that x can be that will make this equation = 0.

To find the zeros, you need to turn 4x^2 -24x+36 into something more simple; change the quadratic equation into something with the form (x+a)(x+b), where as usual, a and b are usually real numbers in beginning classes. This means they can be postive, negative, or fractions.

Have you done something like this before?

If you haven't we can give you hints.

Please, anyone else who knows how to do this, please refrain from giving the answer until PC2 has tried himself (i.e., has posted the correct answer).
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