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dividing fractions
- Thread starter dbf064
- Start date
rule: a/b / (c/d) = ad / (bc)dbf064 said:How do I do (x/y)/z? thanks
your (x/y)/z is really (x/y) / (z/1), so x*1 / (y*z) or x / (yz)
Hello, dbf064!
Here are two ways to handle this . . .
Multiply top and bottom by the LCD, \(\displaystyle y\): . \(\displaystyle \L\frac{y\,\cdot\,\frac{x}{y}}{y\,\cdot\,\frac{z}{1}}\)
And we get: . \(\displaystyle \L\frac{x}{yz}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Do you remember "Invert and multiply"?
We are given: . \(\displaystyle \L\frac{\frac{x}{y}}{z}\) . . This means: \(\displaystyle \L\frac{x}{y}\) is divided by \(\displaystyle z\).
So we have: . \(\displaystyle \L\frac{x}{y}\,\div\,z\;=\;\frac{x}{y}\,\times\,\frac{1}{z}\;=\;\frac{x}{yz}\)
Here are two ways to handle this . . .
We have a complex fraction, \(\displaystyle \:\L\frac{\frac{x}{y}}{\frac{z}{1}}\) . . . one with more than two "levels".How do I do (x/y)/z? thanks
Multiply top and bottom by the LCD, \(\displaystyle y\): . \(\displaystyle \L\frac{y\,\cdot\,\frac{x}{y}}{y\,\cdot\,\frac{z}{1}}\)
And we get: . \(\displaystyle \L\frac{x}{yz}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Do you remember "Invert and multiply"?
We are given: . \(\displaystyle \L\frac{\frac{x}{y}}{z}\) . . This means: \(\displaystyle \L\frac{x}{y}\) is divided by \(\displaystyle z\).
So we have: . \(\displaystyle \L\frac{x}{y}\,\div\,z\;=\;\frac{x}{y}\,\times\,\frac{1}{z}\;=\;\frac{x}{yz}\)