Simplify This Fraction

**MathClown** · October 15th 2012, 01:52 AM

2a/9*6/a^2 = 4/3a. I've played around with the figures but could never derive the answer in bold. How is it done?

Thanks guys.

**BobP** · October 15th 2012, 04:16 AM

Begin by removing the ambiguity on the LHS. What does it actually mean ?

Consider for example the expression 2*6/4*3.

Does this mean 12/4*3 = 3*3 = 9 ?

Or does it mean 12/12 = 1 ?

Remove the ambiguity with the use of brackets.

For example (2*6)/(4*3) = 12/12 = 1.

Operations within brackets are carried out first.

**alane1994** · October 15th 2012, 05:46 AM

Hello MathClown,
Here is how you get the answer in bold.

$\frac{2a}{9}\times\frac{6}{a^2}=\frac{4}{3a}$

$2a\times6=12a$

$9\times{a^2}=9a^2$

So that would make it.

$\frac{12a}{9a^2}$

Then you reduce.

$\frac{4}{3a}$

Does this make sense?

**Soroban** · October 15th 2012, 06:11 AM

Hello, MathClown!

$\text{Verify: }\: \frac{2a}{9}\cdot\frac{6}{a^2} \:=\:\frac{4}{3a}$

I assume you know how to multiply fractions . . .

. . . $\frac{2a}{9} \cdot \frac{6}{a^2} \;=\;\frac{2a\cdot6}{9\cdot a^2} \;=\; \frac{12a}{9a^2}$

. . . and how to reduce fractions.

. . . $\frac{12a}{9a^2} \;=\;\frac{3a\cdot 4}{3a\cdot 3a} \;=\; \frac{\rlap{//}3a\cdot4}{\rlap{//}3a\cdot 3a} \;=\;\frac{4}{3a}$

So exactly where is your difficulty?

**alane1994** · October 15th 2012, 09:28 AM

I assumed that he knew how to reduce...
What I don't get is how you could have problems with this sort of problem. This is some of the most basic algebraic fractions work I can think of. But I guess some people have more difficulty with it than others.
If you have any more troubles with this MathClown. Feel free to message me or just continue on in this thread.

**MathClown** · October 16th 2012, 02:33 AM

Thanks for your help guys. I'm a complete math newbie. I do okay in other subjects, but basic math... ehhh.

Here's one fraction that needs simplifying. I have worked it out.

15a^2b^4c/10a^4b^2c^2 = 3a^2/2b^2c <----- this is my answer. The book's answer is 3b^2/2a^2c Why isn't it correct MY WAY? Also, why are you cancelling out a^4 by a^2 and b^4 by b^2? In the question, a^4 is the denominator and b^4 is the numerator -- why are they both equally being cancelled out? Does it not matter whether they are numerators or denominators... you just cancel out the higher exponent by the lower one, is that how it works?

Thanks guys!

**alane1994** · October 16th 2012, 04:58 AM

$\frac{15a^2b^4c}{10a^4b^2c^2}=\frac{3a^2}{2b^2c}$

Honestly, I'm not sure why this is. I can think of one way but it doesn't make sense in terms of mathematics.

My thoughts.
I think maybe the fourth power of the a in the denominator cancels out the fourth power of the b in the numerator. That gets rid of those two terms...? Then the c on top cancels out the square on the c in the denominator. Divide by 5. This doesn't make any sense to me, but it is the only way that I can think of to get the "correct" answer. Are you sure that that is the correct answer?

**MathClown** · October 16th 2012, 07:27 AM

Originally Posted by alane1994

$\frac{15a^2b^4c}{10a^4b^2c^2}=\frac{3a^2}{2b^2c}$

Honestly, I'm not sure why this is. I can think of one way but it doesn't make sense in terms of mathematics.

My thoughts.
I think maybe the fourth power of the a in the denominator cancels out the fourth power of the b in the numerator. That gets rid of those two terms...? Then the c on top cancels out the square on the c in the denominator. Divide by 5. This doesn't make any sense to me, but it is the only way that I can think of to get the "correct" answer. Are you sure that that is the correct answer?

Positive. I just double-checked the book's answer. My book is frustrating because it doesn't show the steps on how the simplification was derived; it merely shows the answer. So, alane1994, you agree that my answer is correct, and my book's answer is wrong?

**BobP** · October 16th 2012, 09:04 AM

As an example.

$a^{5} = a\times a\times a\times a\times a,$

i.e $a^{5}$ is $a$ multiplied by itself 5 times.

Similarly $a^{3} = a\times a\times a.$

So, $\frac{a^{5}}{a^{3}} = \frac{a\times a\times a\times a\times a}{a\times a\times a}=a\times a = a^{2},$

and $\frac{a^{3}}{a^{5}} = \frac{a\times a\times a}{a\times a\times a\times a\times a} = \frac{1}{a\times a} = \frac{1}{a^{2}}.$

In both cases three a's on the top line cancel with three a's on the bottom line. In the first case we are left with two a's on the top line, in the second case with two a's on the bottom line.

And BTW, the book answer is correct.

**uperkurk** · October 16th 2012, 09:30 AM

Originally Posted by alane1994

I assumed that he knew how to reduce...
What I don't get is how you could have problems with this sort of problem. This is some of the most basic algebraic fractions work I can think of. But I guess some people have more difficulty with it than others.
If you have any more troubles with this MathClown. Feel free to message me or just continue on in this thread.

Awesome, what a way to crap on someone who is trying their best to understand a problem. I'm glad to see you helped him though but that comment really wasn't needed. I too struggle a lot with fractions and using them with algebra. Just because you understand a particular subject with ease others may not.

**alane1994** · October 16th 2012, 10:04 AM

I apologize if that came off as a bit harsh. You said that you have problems with them as well? If you have any questions, message me or start a thread! I would be happy to help you with any of your problems!

**alane1994** · October 16th 2012, 10:10 AM

Originally Posted by BobP

As an example.

$a^{5} = a\times a\times a\times a\times a,$

i.e $a^{5}$ is $a$ multiplied by itself 5 times.

Similarly $a^{3} = a\times a\times a.$

So, $\frac{a^{5}}{a^{3}} = \frac{a\times a\times a\times a\times a}{a\times a\times a}=a\times a = a^{2},$

and $\frac{a^{3}}{a^{5}} = \frac{a\times a\times a}{a\times a\times a\times a\times a} = \frac{1}{a\times a} = \frac{1}{a^{2}}.$

In both cases three a's on the top line cancel with three a's on the bottom line. In the first case we are left with two a's on the top line, in the second case with two a's on the bottom line.

And BTW, the book answer is correct.

But to look at it, it appears based on the answer that...

$\frac{x^3}{x^7}$

$\frac{x \times x \times x\times x}{x \times x \times x\times x \times x \times x\times x}$

Which would equal...

$x^3$

The answer doesn't make sense to me either. So if you know how to work this out, show some steps please?

**MathClown** · October 16th 2012, 11:04 PM

Originally Posted by BobP

As an example.

$a^{5} = a\times a\times a\times a\times a,$

i.e $a^{5}$ is $a$ multiplied by itself 5 times.

Similarly $a^{3} = a\times a\times a.$

So, $\frac{a^{5}}{a^{3}} = \frac{a\times a\times a\times a\times a}{a\times a\times a}=a\times a = a^{2},$

and $\frac{a^{3}}{a^{5}} = \frac{a\times a\times a}{a\times a\times a\times a\times a} = \frac{1}{a\times a} = \frac{1}{a^{2}}.$

In both cases three a's on the top line cancel with three a's on the bottom line. In the first case we are left with two a's on the top line, in the second case with two a's on the bottom line.

And BTW, the book answer is correct.

Mate, I understand where you're coming from, but I don't see how I can apply this to the problem I posted in post #6. I don't understand why my answer in that post is wrong and the book's answer is right. Could you run me through the process of how the book derived that figure?

Many thanks!

**BobP** · October 17th 2012, 12:49 AM

Dealing with the x^3/x^7 problem first, begin by seeing that the fraction can be written as $\frac{1\times a^{3}}{1\times a^{7}},$ which can be written in the expanded form

$\frac{1\times a\times a\times a}{1\times a\times a\times a\times a\times a\times a\times a}.$

Now cancel an $a$ on the top line with an $a$ on the bottom line as many (three) times as you can. You are left with

$\frac{1}{1\times a\times a\times a\times a}=\frac{1}{a^{4}}.$

It isn't necessary to write down all of these $a$ 's, simply subtract the smaller index (number) from the larger one. In this case there will be a surplus of 4 on the bottom line. Had the expression been a^7/a^3 the result would be a^4, since now there is a surplus of 4 on the top line.

Similarly, for example,

$\frac{a^{15}}{a^{8}}=a^{7}, \quad\quad \frac{b^{7}}{b^{23}}=\frac{1}{b^{16}}.$

There could be more than one variable within the fraction. In this case simply deal with each variable separately.

So, with your book example there will be a surplus of two a'2 on the bottom line, two b'2 on the top line and one c on the bottom line.

**kalwin** · October 18th 2012, 11:10 PM

Solve :-
2a / 9 x 6 / a ^ 2 = 4 / 3 a
By cross multiplication method :-
= 4 a / 3 a ^ 2= 4 / 3 a
= 12 a ^ 2 = 12 a^2
= 1 (answer)

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