Complex fractions HELP!
- Thread starter lrou98
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Are you able to combine two fractions, like 1/2+1/3 ?
The process is the same, for combining these two fractions:
1/3 - 1/x
Can you combine these?
That is correct, for combining 1/9 - 1/x^2
Can you do it?
Once you have combined the top into a single ratio, and combined the bottom into a single ratio, you will have one ratio divided by another. You may then divide those by changing to multiplication, using this rule:
a/b divided by c/d
is the same as
a/b times d/c
Factor the difference of squares, and you will see lots of cancellations.
Specific questions on this? :cool:
There are other methods; do you remember seeing any examples? (I don't know what your class is doing.)
D
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Do you realize that the denominator is of the form a2 - b2?
a2 - b2 can be factorized into (a+b)(a-b).
That can make your life simpler.....
\(\displaystyle \dfrac{\dfrac{1}{3} \ - \ \dfrac{1}{x}}{\dfrac{1}{9} \ - \ \dfrac{1}{x^2}} \ = \)
\(\displaystyle \dfrac{\bigg(\dfrac{9x^2}{1}\bigg)\bigg(\dfrac{1}{3}\bigg) \ - \ \bigg(\dfrac{9x^2}{1}\bigg)\bigg(\dfrac{1}{x}\bigg)}{\bigg(\dfrac{9x^2}{1}\bigg) \bigg(\dfrac{1}{9} \bigg) \ - \ \bigg(\dfrac{9x^2}{1}\bigg)\bigg(\dfrac{1}{x^2} \bigg)} \ = \)
\(\displaystyle \dfrac{3x^2 \ - \ 9x}{x^2 \ - \ 9} \ = \ \)
Can you finish that?
And Irou98, you should see that my method uses the least common multiple as you stated, 9x^2.
And it cuts right away to the heart of the matter by eliminating all of the "smaller-image"fractions.
It reduces the work of four lines of the complex fraction down to two lines having just one fraction bar.
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Here is a little trick I have found helpful sometimes:
\(\displaystyle Let\ y = \dfrac{\dfrac{1}{3} - \dfrac{1}{x}}{\dfrac{1}{9} - \dfrac{1}{x^2}}.\) Now I have an equation to work with.
Denis's rule is GET RID OF FRACTIONS. So
\(\displaystyle y * \left(\dfrac{1}{9} - \dfrac{1}{x^2}\right) = \left(\dfrac{1}{9} - \dfrac{1}{x^2}\right) * \left(\dfrac{\dfrac{1}{3} - \dfrac{1}{x}}{\dfrac{1}{9} - \dfrac{1}{x^2}}\right).\) I multiply both sides of the equation by the denominator of the BIG fraction.
This simplies to: \(\displaystyle y * \left(\dfrac{1}{9} - \dfrac{1}{x^2}\right) = \dfrac{1}{3} - \dfrac{1}{x}.\)
THE FRACTIONS ARE NOT SO AWFUL LOOKING, BUT I STILL HAVE LOTS OF THEM. DENIS SAYS GET RID OF THEM. So
\(\displaystyle y * 9x^2 * \left(\dfrac{1}{9} - \dfrac{1}{x^2}\right) = 9x^2 * \left(\dfrac{1}{3} - \dfrac{1}{x}\right).\)
I multiply both sides of the equation by the denominators of the fractions on the left side of the equation to eliminate those fractions.
The result of this last operation is what? If you still have fractions somewhere, you now know how to get rid of them.
Once you have eliminated all fractions, simplify if possible and solve for y. This method is slow, but sure.
Sorry that I mistakenly burdened you with a rule that does not belong to you. After this, I shall make sure to call it Jeff's rule.When/where did I say that? That's just common sense, isn't it