I need some help with my math (algebra) ! :( Can someone help me please?

[WarDemon666]

Simplify the expressions (no negative powers, use exponent rules)

a)
4x^-3y^-5
------------------
6x^-4y^3


b)
( 3x^-2 )^2
( ----------- )
( 4^-1y^3 )

each ^ is an exponent, like 4x to the negative 3 and y to the negative 5

If possible, could u guys give me some details on how to get to the answer?

Ive been working on my math all day and just cant seem to understand

Thanks in advance!!
Carl

 

[GtPrOjEcTX]

yeah, I was wrong. time to sleep

 

[AgaBoogaBoo]

do them step by step

4x^-3
-----------
6x^-4

4/6 = 2/3, correct?

so, now we have

2x^-3
-------
3x^-4

which is the same as 2(1/x^3) over 3(1/x^4)

diving is the same as multiply after flipping the equation, rgiht? like 2 divded by 3/5 is same as 2 times 5/3

So you do 2/x^3 times x^4/3
so it becomes 2x^4/3x^3

So 2x/3 or 1.5X

 

[AgaBoogaBoo]

Do something similar for the other parts, that's not the entire answer, do the rest of that one and we'll check your answer for yo u

 

[WarDemon666]

umm for the second part i got this and got stuck:

y^-5
--------
y^3


= (1)
(----)
(y^5)
---------------------
(1)
(---)
(y^3)



= (1) (y^3)
(--) X (-----)
(y^5) (1)




= (y^3)
------
(y^5)

??

so the whole answer is:

2x(y^3)
-----------
3(y^5)




is this right?!!??

Thanks again!!!

 

[WarDemon666]

About problem b), theres no common factors to start off with, so what should i do?

Thanks again..

Carl

 

[WarDemon666]

Anyone??

 

[DrPizza]

if there are no addition or subtraction signs, a negative exponent simply means it's in the wrong part of the fraction... if it's x^-5 in the denominator, move it to x^5 in the numerator (or vice versa)

Then, cancel as usual. If you have any trouble figuring out the rules once you have positive exponents, change x^5 to x*x*x*x*x. (Or, if you're good with fractions in fractions, change x^-5 to 1/(x*x*x*x*x)

 

[upsciLLion]

Like AgaBooga said, it's much easier to do things step by step rather than trying to cram a bunch of work into one progression to the next line(s).

That is, work with one thing at a time. For instance move things with negative exponents to the opposite side of the vinculum (the bar), then cancel and/or combine x and y terms if you can, then reduce coefficients. Be sure to obey the order of operations too. That can throw your answer off.

At any rate,

a) Move the x^-3 and the y^-6 to the bottom, then move the x^-4 that's on the bottom to the top. Next cancel the x terms so that you have only one x on the top. Combine the y terms to get y^8 on the bottom. Now reduce the 6/4 to 3/2 so you're left with 3/(2y^8).

After understanding the step by step process you should be able to solve b) on your own. Hope this helps!

ups

 

[WarDemon666]

So the answer for a would be:

3
---------
2y^8


Ok makes sense..... im starting to understand

Thanks guys!!!!

 

[Mardeth]

The b one is tough.

My educated guess would be:

(3x^-2)^2
--------------- =
(4^-1)(y^3)

4(3x)^4
------------
y^3

optional:

324x^4
----------
y^3

Now Im not sure if thats right. But if it is that probably the prettiest your going to get...

 

[upsciLLion]

Originally posted by: WarDemon666
So the answer for a would be:

3
---------
2y^8


Ok makes sense..... im starting to understand

Thanks guys!!!!

That's what I got.

For b I got 36/x^4y^3.

 

[Qacer]

Originally posted by: WarDemon666
Simplify the expressions (no negative powers, use exponent rules)

a)
4x^-3y^-5
------------------
6x^-4y^3

Dammit! I thought that meant 4x^-(3y^-5) / 6x^-(4y^3) ... tsk.. tsk.. Mathematica spits out = (2x) / (3y^8)
.. make it clearer... is the denominator of the 2nd part ( 4^-1y^3 ) = 4^-(y^3) ?