Find Products and simplify.

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Krilian

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Oct 3, 2009
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I'm trying to teach myself how to do these problems, so thank you for explaining how to me.

2z^3 / 3 * 21/z^2


3p - 3 / p * 3p^2 / 4p - 4

9x^13 / 4x^6 / 81x / 36x^5
 


\(\displaystyle \frac{2z^3}{3} \cdot \frac{21}{z^2}\)

The rule for multiplying two ratios is (numerator times numerator) over (denominator times denominator).

Symbolically, the rule looks like this:

a/b * c/d = (ac)/(bd)

However, before we multiply two ratios, we should inspect to see whether or not there are common factors above and below because, if there are, they can be canceled before we multiply the ratios.

Here's the factored form of the two given ratios.

\(\displaystyle \frac{2 \cdot z \cdot z \cdot z}{3} \cdot \frac{3 \cdot 7}{z \cdot z}\)

Can you inspect these factors and "see" that both 3 and z^2 are common factor above and below?

After we cancel these common factors, we get the following.

\(\displaystyle \frac{2 \cdot z}{1} \cdot \frac{7}{1}\)

Now we can use the rule for multiplying two ratios.

Numerator times number is (2z)(7).

Denominator times denominator is (1)(1).

So, the product is 14z/1, which we can simply write as 14z.

Questions, or ready for the next one?

 


Crud. I mistakenly edited my first post, instead of appending the above post. You'll have to read up, instead of down. :roll:

Oh well. We remember what I asked, and your answer is still visible.

(I feel like blechh, today. :| )

 
Re:

mmm4444bot said:
Underneath each of your typed expressions (shown in red), I have provided a guess as to what I think you mean.

Please confirm or correct my guesses, and we'll go from there.

2z^3 / 3 * 21/z^2

\(\displaystyle \frac{2z^3}{3} \cdot \frac{21}{z^2}\)

3p - 3 / p * 3p^2 / 4p - 4

\(\displaystyle \frac{3p \;-\; 3}{p} \cdot \frac{3p^2}{4p \;-\; 4}\)

9x^13 / 4x^6 / 81x / 36x^5

\(\displaystyle \frac{9x^{13}}{4x^6} \div \frac{81x}{36x^5}\)
Click to expand...

Remember that multiplication does not require that we use "like" denominators. Therefore, the multiplications follows normal rules of exponents. Remember that that when dividing exponents of like base, subtract the powers and retain the base.

For example,

\(\displaystyle \frac{2z^3}{3} \cdot \frac{21}{z^2}\) = \(\displaystyle \frac{2z^3 \cdot 21}{3 \cdot z^2}\) = \(\displaystyle \frac{2 \cdot 21 \cdot z^3}{3 \cdot z^2}\) = \(\displaystyle \frac{42 \cdot z^3}{3 \cdot z^2}\) =

\(\displaystyle \frac{42}{3} \cdot \frac{z^3}{z^2}\) = \(\displaystyle 14 \cdot z^{(3-2)}\) = 14z
 


Thanks, Barson!

You fortuitously quoted my post at the very moment that I was stupidly deleting it. :D

 
barson90 said:
3p - 3 / p * 3p^2 / 4p - 4

\(\displaystyle \frac{3p \;-\; 3}{p} \cdot \frac{3p^2}{4p \;-\; 4}\)
Click to expand...


Okay, first a comment about proper notation when typing algebraic ratios using a keyboard: Use grouping symbols!

To clearly show what's in numerators versus denominators, we enclose each in parentheses.

Here's how to properly type the expression.

(3p - 3)/p * (3p^2)/(4p - 4)

So, the rule is the same, but, in this case, we need to do some factoring before we can "see" potential common factor(s) to cancel.

\(\displaystyle \frac{3(p \;-\; 1)}{p} \cdot \frac{3 \cdot p \cdot p}{4(p \;-\; 1)}\)

Can you continue, from here?

 
barson90 said:
9x^13 / 4x^6 / 81x / 36x^5

\(\displaystyle \frac{9x^{13}}{4x^6} \div \frac{81x}{36x^5}\)
Click to expand...


Again, I'm starting by showing you how to properly type this compound ratio.

[(9x^13)/(4x^6)] / [(81x)/(36x^5)]

Now, here's a new rule for you to memorize.

Dividing by a fraction is the same as multiplying by its reciprocal, instead.

Symbolically, it looks like this.

(a/b)/(c/d) = (a/b)*(d/c)

So, we have the following.

\(\displaystyle \frac{9x^{13}}{4x^6} \cdot \frac{36x^5}{81x}\)

Finish it by first cancelling common factors and then multiplying, as is the case in the first two exercises.

 
Krilian said:
Or is the answer (x^11) YUP!
Click to expand...


In future posts, please start by showing whatever work that you can.

This will preclude us from typing a bunch of stuff that you already know.

 
Thank you very much by helping me. I wish I could type my problems in the same bold, correct format you guys do.

Thank you
 
Krilian said:
do you have any advice on how I can get used to doing these problems.
Click to expand...
PRACTICE! However, you're doing pretty well...
 
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