Xtream Help

Note, DS, that we have the law of exponents that says  aⁿ / a^m   = a ^n-m

So.....look at the fraction....we have

7 * ( a² / a⁵)     and using this law, we have

7 * (a ^2-5 )  =   7 * a^-3

And remember that  a^-n   = 1 / aⁿ

So.......

7 * a^-3  =   7 * ( 1 / a³)  = (7 / a³ )

So...our answer should be

a²b - (7/a³)     

Does that help ???

It does, but it is confusing. And the answer is not there in the homework. I will shoot the photo, and you would see that the answer isn't there, and it will be my proof and evidence that the answer is not there. So, do I shoot the photo?

Isn't my answer the same thing as your second "possible" answer ???

I'm confused !!!    ????????

Yes it is. And it is incorrect, I guess. I will shoot the photo.

Only help on T.

I may need help on the ones of subtracting. Can you teach me how to do it?

OK.....if we have

[a²b - 7a²] / a⁵       we can factor out an a²  at the top and we have

[(a²) (b - 7)] / a⁵   =

[b - 7 ] / a³ 

I have a question. In the first step, why do you cancel out  $$a^2?$$

Hi Dragon  

$$a^2b-7\frac{a^2}{a^5}\\\\ a^2b-\frac{7}{1}\times\frac{a^2}{a^5}\\\\ a^2b-\frac{7\times a^2}{1\times a^5}\\\\ a^2b-\frac{7a^2}{a^5}\\\\ a^2b-\frac{7aa}{aaaaa}\\\\ a^2b-\frac{7\not{a}\not{a}}{\not{a}\not{a}aaa}\qquad \mbox{Note: aa cancels out}\\\\ a^2b-\frac{7}{aaa}\\\\ a^2b-\frac{7}{a^3}\\\\$$

I am more confused.

Dragon, you wrote that before you even had time to think about what I had written!!!

Now go back and think about it properly!

May someone explain how you canceled out aa???

And can someone help me figure out to do in the following problem please?:

$$\dfrac{6a^2-30a+36}{4a-12}$$  

How do I start off, and can you give an example of how to do it?

=6(a^2-5a+6)/4(a-3)=6(a-2)(a-3)/4(a-3)=3/2*(a-2)=3*a/2-3

Melody, you're answer is not on the sheet.

It was your very fist MathWay question!  the answer had a blue tick!

[ 6a² -30a + 36 ] / [4a - 12 ]   factor out the common factor of 6 at the top and 4 at the bottom

(6/4) [a² - 5a + 6 ] / [a - 3]   factoring the trinomial, we have

(6/4) [(a -3)(a-2)] / (a-3)      "canceling the "(a-3)'s "  and reducing (6/4) we have

(3/2)(a - 2) or   (3a - 6) / 2    ...whichever you prefer !!!

Oh! I get the answer now! Melody, how do you know it is my first MathWay question? This is my third, but I didn't post it here...

Can you guys help on the following?

$$\dfrac{3a^3(16-a^2)}{12a^6(a^2-9a+20)}$$

I got to the part:

$$\dfrac{1(16-aa)}{4a^3(a^2-9a+20)}$$  

I don't know what to do next. Can you pleas tell me what to do next?

May anyone help me here!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!??!?!?!??!!?!

3a³(16- a²) / [12a⁶ (a² - 9a + 20)]   factoring everything, we have

3a³ (4 - a)(4+a) /[ 12a⁶ (a - 4)(a-5)] =

-(1/4a³ )(a-4)(a+4)/[(a-4)(a-5)] =   (we can factor a "-" out of (4 - a) = - (a - 4))

-(1/4a³)(a+4)/(a-5) = (5-a)(4+a)/(4a³)

Thank you! I need to understand what to do for other similar questions like that one. Can you please tell or explain what to do in questions like these?

thanks. I would give you hundreds and hundreds of points CPhill.

I don't have any "magic" formula for doing these, DS......it's mainly just factoring and "cancelling"......plus a few algebraic "tricks".......

Much of this is WAY OVER your head Dragon.  Your teacher cannot seriously expect you to do it!!

I know. The teacher said, " try your best. I will tell you how on Friday" 

so, I try to make the best work I can. Ok. I know. It is not my level, but thanks for helping.

you are welcome Dragon