my problem is
1/2a3 + 3/4ab2 + c/8a2b2
the red numbers are exponets
does any one know how to do this?
thanks
you need to find the least common denominator of all three fractions. $\displaystyle 8a^3b^2$.
now you can multiply through each fraction with LCD. use the same cancellation rules for regular fractions then wright your numerators over the LCD.
Since you are not solving this equation all you have to do is add these fractions together by first getting the same common denominator and only then can you add them together.
Example with numbers:
$\displaystyle \frac{1}{6}+\frac{1}{3}$
$\displaystyle \frac{1}{6}+\frac{1}{3}*\frac{2}{2}$
$\displaystyle \frac{1}{6}+\frac{2}{6}= \frac{3}{6}$
Posting this way would be ok: 1 / (2a^3) + 3 / (4a b^2) + c / (8a^2 b^2)
The denominators are now CLEAR: OK?
WHY did you say "equal denominators"?
You can start by multiplying through by 2a; result:
1 / (a^2) + 3 / (2b^2) + c / (4a b^2)
Can you "see" that the LCD is (4 a^2 b^2) ?
I labeled it equal denominators because thats what the lesson called it. Thanks for showing a diffrent way to write it. I still dont get how you got 4a^2b^2 though because if it was going with the largest numbers itd be 8a^3b^2 or,if going with the smaller numbers,itd be 2a^3
Looks like you didn't "follow" what I did to simplify the original:
1 / (2a^3) + 3 / (4a b^2) + c / (8a^2 b^2)
to this:
1 / (a^2) + 3 / (2b^2) + c / (4a b^2)
by multiplying each term by 2a;
do you "see" that 2a is common to all 3 denominators?
So:
1 / (2a^3) * 2a = 1 / a^2
3 / (4a b^2) * 2a = 3 / (2b^2)
c / (8a^2 b^2) * 2a = c / (4a b^2)
It is not a must to do this simplification: it simply makes it EASIER,
since you are now working with expression:
1 / (a^2) + 3 / (2b^2) + c / (4a b^2)
and with LCD:
4a^2 b^2
Let's do it then; we have 1 / (a^2) + 3 / (2b^2) + c / (4a b^2), with LCD 4a^2 b^2
Each term is to be multiplied by the LCD:
1 / a^2 * 4a^2b^2 = 4b^2
3 / (2b^2) * 4a^2b^2 = 6a^2
c / (4ab^2) * 4a^2b^2 = ac
So you now have:
(4b^2 + 6a^2 + ac) / (4a^2 b^2)
Now, to reverse the multiplication by 2a we did to simplify:
(4b^2 + 6a^2 + ac) / [2a(4a^2 b^2)]
or:
(4b^2 + 6a^2 + ac) / (8a^3 b^2)
You can prove to yourself that this is correct by assigning any
values to a,b,c (like a=2,b=3,c=4) and substituting these in the
original expression, and in the simplified expression: both must be equal.
Much easier if 1 is assigned to all variables: a = b = c = 1; then:
original expression = 1 / (2a^3) + 3 / (4a b^2) + c / (8a^2 b^2)
= 1/2 + 3/4 + 1/8 = 4/8 + 6/8 + 1/8 = 11/8
and simplified expression = (4b^2 + 6a^2 + ac) / (8a^3 b^2)
= (4 + 6 + 1) / (8) = 11/8
Got that?
@Wilmer & Krae, To both of you.
Part of the problem with this thread is the difficulty in read the sloppy notation. Answering with equally hard to read notation is not helpful in my view. So why not learn to post in symbols? You can use LaTeX tags http://www.mathhelpforum.com/math-he...ial-19060.html
Here is an example:
[TEX]\frac{1}{2a^3}+\frac{3}{4ab^2}+\frac{c}{8a^2b^2}[/TEX]
gives $\displaystyle \frac{1}{2a^3}+\frac{3}{4ab^2}+\frac{c}{8a^2b^2}$
That is a bit more typing but the payoff is enormous. It is so helpful to be able to easily read and understand. It is easy to learn LaTeX. The easier it gets the more you use it.
$\displaystyle \frac{1}{2a^3}+\frac{3}{4ab^2}+\frac{c}{8a^2b^2}$
We see that we require $\displaystyle 8,\;\;a^3,\;\;b^2$
in the denominators.
Multiply each fraction by 1, in such a way as to get the required denominator.
$\displaystyle \frac{1}{2a^3}\left(\frac{4b^2}{4b^2}\right)+\frac {3}{4ab^2}\left(\frac{2a^2}{2a^2}\right)+\frac{c}{ 8a^2b^2}\left(\frac{a}{a}\right)$