Please Help!!

[0411G053A]

I need some help with this problem:

27a^x+y b^x-y
Simplify: ---------------
-3a^x-y b^x+y

I am not sure of how to simplify this.

[Guy Desautels]

Before I attemp this: Is there a space between y and b^x; if yes, then
should there not be some operation or parentheses?

(27a^x+y) (b^x-y)
> Simplify: ------------------
> (-3a^x-y) (b^x+y)

(27a^x) + (yb^x-y)
> Simplify: -------------------
> (-3a^x) - (b^x+y)

27a^x + yb^x - y
> Simplify: ------------------
> -3a^x -yb^x +y

Which one above???

--
Guy Desautels

[GWinstonHW]

In article <4t05tt$1f...@ausnews.austin.ibm.com>, 0411G053A@ writes:

>
> 27a^x+y b^x-y
>Simplify: ---------------
> -3a^x-y b^x+y

I'm assuming a is raised to power (x+y). Brackets would have been useful.

At school, you learn (a^x) / (a^y) = a^(x-y). You can apply the same
principle to this question.

For example, a^(x+y) / a^(x-y) = a^([x+y]-[x-y]) = a^(2y)

The whole equation simplifies to -9 a^(2y) b^(-2y), but since n^(-a) =
1/(n^a), this can rewritten as -9 a^(2y) / b^(-2y).

Gavin

[Sheldon Ackerman]

GWinstonHW (gwins...@aol.com) wrote:

: The whole equation simplifies to -9 a^(2y) b^(-2y), but since n^(-a) =

: 1/(n^a), this can rewritten as -9 a^(2y) / b^(-2y).

^^^^^^
I'm sure it's a typo on your part, but the part I underlined with ^ above,
should read:

b^(2y)

--
Sheldon Ackerman.......http://www.dorsai.org/~ackerman/
acke...@dorsai.dorsai.org

[GWinstonHW]

In article <4t132j$7...@amanda.dorsai.org>, acke...@dorsai.org (Sheldon
Ackerman) writes:

>I'm sure it's a typo on your part, but the part I underlined with ^
above,
>should read:
>
>b^(2y)

Yes, it was a typo. Thanks for the correction.

Gavin

[0411G053A]

In <4t0qq8$c...@newsbf02.news.aol.com>, gwins...@aol.com (GWinstonHW) writes:
>In article <4t05tt$1f...@ausnews.austin.ibm.com>, 0411G053A@ writes:
>
>>
>> 27a^x+y b^x-y
>>Simplify: ---------------
>> -3a^x-y b^x+y
>
>I'm assuming a is raised to power (x+y). Brackets would have been useful.
>
>At school, you learn (a^x) / (a^y) = a^(x-y). You can apply the same
>principle to this question.
>
>For example, a^(x+y) / a^(x-y) = a^([x+y]-[x-y]) = a^(2y)
>

>The whole equation simplifies to -9 a^(2y) b^(-2y), but since n^(-a) =
>1/(n^a), this can rewritten as -9 a^(2y) / b^(-2y).
>

>Gavin

Thanks,
This helps very much. I will be sure to stay here for the help. :)>

[gars...@pipeline.com]

Please, when you do, remember the brackets.

My guess as to what you meant was way off.


-- Gary Simon
>

[Conn E. Smith]

gars...@pipeline.com wrote:
>
> In article <4t5kti$18...@ausnews.austin.ibm.com>, 0411G053A@ writes:
>
> >In <4t0qq8$c...@newsbf02.news.aol.com>, gwins...@aol.com (GWinstonHW)
> writes:
> >>In article <4t05tt$1f...@ausnews.austin.ibm.com>, 0411G053A@ writes:
> >>
> >>>
> >>> 27a^x+y b^x-y
> >>>Simplify: ---------------
> >>> -3a^x-y b^x+y
> >>
> >>I'm assuming a is raised to power (x+y). Brackets would have been useful.

[ I cut out balance of this solution ]

To whomever is interested:

In the above expression the (x + y) is NOT a power. The (x + y ) should
be referred to by one of its proper names: EXPONENT or if you prefer,
index would also be acceptable.

In the expression 3^2, the 3 is the base, the 2 is the exponent and
the whole thing ( 3^2 ) is properly called the power. ie It would be
correct to say that the second power of 3 is 9.