# Can someone give me some help with this algebra?

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From Lial's Beginning algebra:

Why is this true?

8a(to the 15th) b(to the first) =

16a(to the 4) b(to the 6th)

a(to the 11th)

2b(to the 5th)

It's kinda hard to do the exponents, but I hope you get the idea... Where did the 2b come from? The other stuff, I get. Just not the 2b.

Thanks, and please excuse my pathetic algebra skills. :001_huh:

Jackie

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Okay, I'll take a stab at this.

8a(to the 15th) b(to the first)

16a(to the 4) b(to the 6th)

by expanding some of the terms, we see that this is the same as

8a(to the 4th)a(to the 11th)b(to the first)

8(2)a(to the 4th)b(to the first)b(to the fifth)

canceling the 8s, the a(to the 4th) terms, and the b(to the first) terms leaves

a(to the 11th)

2b(to the fifth)

which is what you stated originally.

Regards,

Kareni

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Hi Jackie,

I was able to get your answer --> a(11th)/2b(5th). The best way to explain is to break the problem up as follows:

--> (8/16)(a(15)/a(4))(b(1)/b(6))

then I reduce,

--> (1/2)(a(11)/1)(1/b(5))

then I multiply the numerators to get --> a(11th)

and then I multiply the denominators to get --> 2b(5th)

If this doesn't work for you, another way to solve it is to actually show the representation of the exponents and then simplify:

(8)(aaaaaaaaaaaaaaa)(b) --> aaaaaaaaaaa --> a(11th)

-------------------------- -------------- -------

(16)(aaaa) (bbbbbb) (2) (bbbbb) 2(b(5th))

It is hard to show typing :tongue_smilie::tongue_smilie:, but I do hope this helps you ... good luck!

Terri

Edited by Harrison_B
The lines were not lined up.
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Don't think of the 2 and the b as being together. If you mentally separate the 8/16 from the a^15 b/a^4 b^6, you end up with 1/2 when you reduce the 8/16 to lowest terms. You dont need to show the 1 on top in your answer because it would be like 1 times a^11. The 2 stays on the bottom with the b^5.

That's how I understand it in my head! Lol! Hope this helps!

Edited by Afton
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My mind does it this way (Some days my explanations make so much sense to my kids--other days they fail to connect. If you brain "sees" this problem like I do, great. If not...ignore it. LOL!):

8 a^15 b^1=

16 a^4 b^6

Because of the way exponents work, the exponents in the bottom of an equation can be made negative and be brought into the top of the equation. Thus:

1

b^6

equals

b^-6

In other words, you can write this two ways. If you want to, you can put b to the sixth in the top of the fraction, but you have to switch the sign to negative. If you keep it in the bottom of the fraction, it stays positive. This helps many times when simplifying a problem since you can pull those exponents up or send them below the line as long as you remember to change the sign each time you move it.

So we can move the exponents in the bottom to the top and we get this:

8 times a^15-4 times b^1-6

16

And so we get:

8/16=1/2 and

a^15-4=a^11 and

b^1-6=b^-5 since a negative exponent is the same as the recipricol, you can write this as:

1

b^5

Which then goes together as:

a^11

2b^5

I proofed this 3 times. I hope I did not make any stupid mistakes. It is nasty trying to do this on the computer, isn't it? LOL! This type of moving exponents helps me "see" this problem so much more clearly. I suppose my kids would tell me I'm nuts, though. Well---it is just another option for you.

Jean

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This is the one that clicks for me. I WAS thinking of the 2 and the b as being together. I think you and Jean are saying the same thing. I just didn't "see" the reduction of the 8/16 becoming 2b (after the a was moved to the numerator).

Thanks ladies.... Now I "see" it, and as always when this happens, it seems so obvious.

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