Question

how to simplify a compound fraction? a drawn example would be stellar

Answer 1

You need to convert the whole number first into a fractional equivalent and then add it to the fraction.

Answer 2

\[3\frac{1}{3}=3\frac{3}{3}+\frac{1}{3}=\frac{9}{3}+\frac{1}{3}=\frac{10}{3}\]

Answer 3

Do you have any specific ones you want worked through?

Answer 4

yes hold on let me write it

Answer 5

Keep in mind that the example I gave is the equivalent of \[3+\frac{1}{3}\]

Answer 6

Ok, a written explanation first. Multiply the top and bottom of the outer fraction by the LCD of all the inner fractions:

\[\huge\frac{\frac{1}{a} + \frac{1}{c}}{\frac{1}{ad}}\]\[= \huge \frac{\frac{1}{a} + \frac{1}{c}}{\frac{1}{ad}} \cdot \frac{acd}{acd} \]\[= \huge \frac{cd + ad}{c}\]

Answer 7

Hrm.. I may have misunderstood the question.

Answer 8

\[1\div(1-(3/(2+x))) = 60\]

Answer 9

Oh. yes That's what I was talking about.

Answer 10

I'll admit his question is more fun now than a basic compund fraction :)

Answer 11

\[\huge \frac{1}{1-\frac{3}{2+x}}\]
The LCD of the inner fraction is just (2+x)
\[=\huge \frac{1}{1-\frac{3}{2+x}} \cdot \frac{2+x}{2+x}\]
\[=\huge \frac{2+x}{(2+x) - 3}\]

Answer 12

Just be sure to restrict x to \(x \ne -2\)

Answer 13

when you're finished.

Answer 14

So the simplified form is:
\[\frac{x+2}{x -1}; x \ne -2\]

Answer 15

Don't forget the 60 he has on the right polpak.

Answer 16

Oh! Then we can solve it!

Answer 17

this cant be right.. (2+x)/(2+x) -3 = 60? unless the answer is no solution

Answer 18

I was just simplifying the fraction.

Answer 19

right

Answer 20

If there was no equals sign what he did was right. The important point he was trying to stress is the denominator can never = 0.

Answer 21

i got 59x = 62 i feel as if that is wrong lol

Answer 22

\[\frac{x+2}{x-1} = 60; x \ne -2\]\[\implies x+2 = 60(x-1) ; x \notin \{-2,1\}\]\[\implies 59x = 62; x \notin \{-2,1\}\]\[\implies x =\frac{62}{59}; x\notin \{-2,1\} \]
It has a solution.

Answer 23

That's exactly right

Answer 24

To reiterate the 'trick' of simplifying complex fractions. Multiply the fraction by 1 = L/L where L is the LCD of all of the inner fractions.

Answer 25

But be sure to exclude the case where the LCD might equal 0.

Answer 26

\[\frac{1}{1-\frac{3}{2+x}}=60\]
Take the inverse on both sides.
\[1-\frac{3}{2+x}=\frac{1}{60}\]
\[1-\frac{1}{60}=\frac{3}{2+x}\]
Take the inverse of both sides:
\[\frac{1}{1-\frac{1}{60}}=\frac{2+x}{3}\]
Simplify and multiply both sides by 3:
\[\frac{3}{\frac{59}{60}}=2+x\]
\[\frac{60}{59}3-2=x\]
\[\frac{180}{59}-{118}{59}=x\]
\[x=\frac{62}{59}\]

Answer 27

That works too, but I hate flipping fractions ;p

Answer 28

I have a typo. That should be \[\frac{118}{59}\]
Hurray for typos :)

Answer 29

you guys are smart. thanks a lot

Answer 30

Since it was looking for a specific answer, would the exclusion of the -2 matter in this case? Just general curiousity on my part.

Answer 31

I don't claim to be smart. Just experienced. You'll get there with practice :)

Answer 32

Well, it would matter if the only solution you came up with was x = -2. Then you know that the equation has no solutions.

Answer 33

Or if the solution was x = 1. That's also not allowed.

Answer 34

Makes sense. Thanks for clarifying.