Help me PLEASE! ASAP! thanx!

T

tinyniel

New member
Joined
Aug 18, 2005
Messages
8
hey i have a major problem. here it is.

Numerators: x^2y, 4^2p^3, 2xp, 5y

denominators: p^-3, y^-1, x^-1p^2, p^-3x^-2

i'm supposed to add like terms and write the answer with all the variables in the numerator. could someone please explain this to me?
Thanks.
-ally
p.s. they are all being subtracted.

Edit: Please don't use white.
 
No clue. I think you'll have to present the actual problem statement.
 
Please verify that
((x^2)*y)/(p^-3) - (4^2)(p^3)/(y^-1) -
(2xp)/((x^-1)(p^2)) - (5y)/((p^-3)(x^-2))
is what you mean. I don't trust the 4^2.
And lose the cutesy color scheme.
 
Not clear...4 fractions? Like x^(2y) / p^(-3) - fraction - fraction - fraction?

Also VERY difficult to read: post the WHOLE problem, plus in BLACK!

Edit: thanks TK and Gene :)
 
ok, sorry, heres the problem. i wasnt sure how to type the fractions.

x^2*y/p^-3 - 4^2*p^3/y^-1 - 2xp/x^-1*p^2 - 5y/p^-3*x^-2

is that better? :D

-ally
 
tinyniel said:
x^2*y/p^-3 - 4^2*p^3/y^-1 - 2xp/x^-1*p^2 - 5y/p^-3*x^-2
Click to expand...
Do you mean this?

. . .<sup>(x<sup>2</sup>y)</sup>/<sub>(p<sup>-3</sup>)</sub> - <sup>(4<sup>2</sup>p<sup>3</sup>)</sup>/<sub>(y<sup>-1</sup>)</sub> - <sup>(2xp)</sup>/<sub>(x<sup>-1</sup>p<sup>2</sup>)</sub> - <sup>(5y)</sup>/<sub>(p<sup>-3</sup>x<sup>-2</sup>)</sub>

If so, you could start by moving all the negative-power bases to the other sides of their various fraction lines.

Eliz.
 
Eliz:
Sorry, I like her's better than yours. The double "sup"s didn't work.

ally:
Much better but there are still questionable spots.
2xp/x^-1*p^2
equals
2x²p³
but I think you mean
2xp/(x^-1*p^2)
which equals
2x²/p
If you do mean the first, I apologize but more ()s would have made it clear as in
(2xp/x^-1)*p^2

Same with
5y/p^-3*x^-2

Please look at "order of operations" over in the blue section and the last two posts in
http://www.freemathhelp.com/forum/viewtopic.php?t=2

We all want to help, but we also want to be working on the right problem.

One more time :twisted:
 
Whoops, I forgot about the first post. You do mean what I thought.
x^2*y/p^-3 =
x^2*p^3*y

4^2*p^3/y^-1 =
16*p^3*y

2xp/(x^-1*p^2) =
2*x^2/p

5y/(p^-3*x^-2) =
5*x^2*p^3*y

I'm still suspicious of two spots though. At this level of algebra I would bet a nickle that the 4 in 4^2*p^3/y^-1 is an x and it says
x^2*p^3/y^-1 =
x^2*p^3*y
and
2xp/(x^-1*p^2) actually says
2*x*p*y/(x^-1*p^-2) =
2*x^2*p^3*y

If those two things are true you should be able to finish it.
 
Gene said:
Eliz: Sorry, I like her's better than yours. The double "sup"s didn't work.
Click to expand...
Really? It looks fine in Mozilla. I guess the display is browser-dependent...? Sorry!

Eliz.
 
Eliz,
I should have suspected that. WebTV is very poor in translation. I can't even get to any of the math help sections above "Other Features".
--------------------
Gene
 
let me re-write the problem using "sups". here ya go.

x<sup>2</sup>y/p<sup>-3</sup> - 4x<sup>2</sup>p<sup>3</sup>/y<sup>-1</sup> - 2xp/y<sup> -1</sup>p<sup>2</sup> - 5y/p<sup>-3</sup>x<sup>-2</sup>

hope thats better. thanks for all your help.
-ally

p.s. could explain how to do these problems? i'm so lost. my mom doesnt remember it either.
 
The first step would be to move everything with a negative exponent to the other side of the fraction line. For instance:

. . . . .2/x<sup>-3</sup> = 2x<sup>3</sup>

. . . . .3y<sup>-1</sup> = 3/y<sup>1</sup> = 3/y

. . . . .2m<sup>-3</sup>/4n<sup>-2</sup> = 2n<sup>2</sup>/4m<sup>3</sup>

. . . . .-3z<sup>-4</sup> = -3/z<sup>4</sup>

Simplify this part first. Then we can work with the fractions.

Eliz.
 
i dont exactly know how to do that. i know i should, but i dont. i kinda forgot everything over the summer. lol.
 
Well, your 3rd term 2xp / (y^(-1)p^2 is (to me) WRONGLY typed:
should be 2x^2p / (y^(-1)p^(-2); this makes 'em all the same:

x^2 y p^3 - 4 x^2 y p^3 - 2 x^2 y p^3 - 5 x^2 y p^3
= -10x^2yp^3

Agree?!

I feel like commenting on your carelessness...but will not.

REPEATING what others told you:
a / b^-2 = a * b^2

a * b^-2 = a / b^2
 
You must log in or register to reply here.
Top