Can this be simplified further?

[jtw2e2]

Original problem:

I began by attempting to factor and came up with:

[((x+y)(x-y)) / ((x-y)(x+y)) times ((4x-3y)(x-y)) / (2x(x-3y)-y(x+3y))] I hope that was written clearly enough. If not, let me know and I'll try to rewrite as proper fractions.

I canceled and was left with:

((x+y)(4x-3y)) / (2x(x-3y)-y(x+3y))

Is that correct so far? And is it possible to simplify further? Thanks so much for your help.
-Jeff

 

[mmm4444bot]

… came up with:

[((x+y)(x-y)) / ((x-y)(x+y)) * ((4x-3y)(x-y)) / (2x(x-3y)-y(x+3y))] We can use an asterisk as a multiplication symbol.

Some of these factorizations are wrong.

Also, it looks to me like you flipped the righthand ratio. Were you thinking of the rule that division by a fraction is the same as multiplying by its reciprocal? That rule does not apply in this exercise, since the two given ratios are being multiplied.

Here is your factorization for the lefthand ratio:

((x+y)(x-y)) / ((x-y)(x+y))

Your numerator is the same as your denominator!

Here is the correct factorization, for the lefthand ratio:

(x + y)^2 / [(x + y)(x - y)]

Here is your factorization for the righthand ratio:

((4x-3y)(x-y)) / (2x(x-3y)-y(x+3y))

The factorization that you typed in the numerator is correct for the given denominator. The other factorization is not correct.

Here is the correct factorization, for the righthand ratio:

[-(3x + 4y)(x + y)] / [(4x - 3y)(x - y)]

If you had shown your work on the factorizations, perhaps I could have seen where you went wrong.

Putting it all together, we have the following.

(x + y)^2 / [(x + y)(x - y)] * [-(3x + 4y)(x + y)] / [(4x - 3y)(x - y)]

Can you finish simplifying this ?

Cheers ~ Mark

 

[Denis]

[((x+y)(x-y)) / ((x-y)(x+y)) times ((4x-3y)(x-y)) / (2x(x-3y)-y(x+3y))]

Hey Jeff, you're improving fast!
Nice bracketing job; your square brackets are not necessary; could be used (for clarity) like this:

[((x+y)(x-y)) / ((x-y)(x+y))] times [((4x-3y)(x-y)) / (2x(x-3y)-y(x+3y))]

Couple errors:
your 1st numerator (x+y)(x-y) should be (x+y)(x+y) ; ok?

your 2nd denominator (2x(x-3y)-y(x+3y)) should be (-3x-4y)(x+y) ; ok?

Others ok: good job!

Can you finish it now?

EDIT: Mark, yer making me nervous :wink:

 

[jtw2e2]

Re:

...
(x + y)^2 / [(x + y)(x - y)] * [-(3x + 4y)(x + y)] / [(4x - 3y)(x - y)]

Can you finish simplifying this ?

Cheers ~ Mark

[/color]

Mark,

I apologize for not typing out all the steps--I was in a hurry to leave for school. In high school we never really got into factoring. Do you mind if I ask how one would determine that -(3x+4y)(x+y) is the correct factorization? What I mean to say is that I can FOIL and see that it works, but how did you arrive at that factorization?

....

Hope this makes sense. Please let me know if I was unclear.

Addendum...I see why I was struggling. I'll try again :shock:

 

[jtw2e2]

Re: Re:

Addendum...I see why I was struggling. I'll try again :shock:

Final result: (4x-3y) / (-3x-4y)

According to the website to which I submit my homework, this is the correct answer. Thanks for the help Mark and Denis.

 

[Denis]

Re: Re:

Final result: (4x-3y) / (-3x-4y)

YA!
Could be shown as (4x - 3y) / -(3x + 4y) or -[(4x - 3y) / (3x + 4y)]

You were asking how did we factor say 4x^2 - 7xy + 3y^2.
Well, since we were getting (x-y)'s and (x+y)'s from other terms,
then chances were either x+y or x-y would be a factor; always look for short cuts :idea:

 

[Deleted member 4993]

4x^2 - 7xy + 3y^2

= y[sup:1g370y91]2[/sup:1g370y91] * (4 * (x/y)[sup:1g370y91]2[/sup:1g370y91] - 7 * (x/y) + 3)

for clarity substitute

u = x/y

y[sup:1g370y91]2[/sup:1g370y91] * (4 * (x/y)[sup:1g370y91]2[/sup:1g370y91] - 7 * (x/y) + 3)

= y[sup:1g370y91]2[/sup:1g370y91] * (4 * u[sup:1g370y91]2[/sup:1g370y91] - 7 * u + 3)

= y[sup:1g370y91]2[/sup:1g370y91] * (4 * u - 3)( u - 1)

then substitute back

y[sup:1g370y91]2[/sup:1g370y91] * (4 * u - 3)( u - 1)

= y[sup:1g370y91]2[/sup:1g370y91] * [4 * (x/y) - 3][(x/y) - 1]

= y * [4 * (x/y) - 3] * y * [(x/y) - 1]

= [4 * x - 3 * y] * [x - y]