When does the denominator disappear in a common denominator equation?

[craig0303]

hello, I am pretty up on most pre-calculus math but i have a bit of confusion

when you raise a denominator of an equation so all numerators are on the same level, at what point does the denominator disappear and at what point does it stay after this activity?

eg. 4/4+5/2+6/4=4/2 = 4/4+10/4+6/4=8/4

sometimes the denominator stays after this action, and sometimes it dissapears, what is the actual mathematical rule for this? just gets confusing, sometimes you keep all the denominators after you find a common denominator, sometimes you wipe the denominator and just have the numerator afterwards

thanks for the clarification

 

[HallsofIvy]

I'm not at all clear what you mean by "raising" a denominator. Do you just mean getting common denominators? Remember that a fraction indicates a division. If, after adding fractions, the numerator happens to be a multiple of the denominator, as in the case of 8/4, you can go ahead and do the division and write the result as an integer: 8/4= 2. However, you never "wipe the denominator and just have the numerator". 8/4 can never be written as "8".

I suspect your confusion is due to memorizing formulas rather than understanding what you are doing.

 

[craig0303]

I'm not at all clear what you mean by "raising" a denominator. Do you just mean getting common denominators? Remember that a fraction indicates a division. If, after adding fractions, the numerator happens to be a multiple of the denominator, as in the case of 8/4, you can go ahead and do the division and write the result as an integer: 8/4= 2. However, you never "wipe the denominator and just have the numerator". 8/4 can never be written as "8".

I suspect your confusion is due to memorizing formulas rather than understanding what you are doing.

what about 5A/5+7A/10=9A/10 (just a random question, probably doesn't equate)

im just wondering the general rule, all i know is how to do it in each instance, sucks to move on and be missing a hole here and there, just want to know why sometimes the denominator stays and sometimes when the numerators are on equal level the denominator dissapears

just want to know why sometimes the common denominator stays and why sometimes when its become common throughout the equation it then dissapears and your just left with numerators which have been put on common terms

just want to know the universal rule and reason for all this, all i know is how to do them in particular instances, thanks for any elaborate response, very curious.

 

[pyrotix]

Do you understand friendly or basic fractions? like how 3/6=4/8=1/2 ?

Basically once you get an answer to a fraction problem you want to put in lowest terms.
Say you have 8/12. If we express that in terms of primes it is 2*2*2/2/2/3. Two of the twos that are multiplied cancel with 2 that are divided and you end up with 2/3.

Alternately if you have something 15/3=5*3/3. The 3's cancel and you are left with 5.

 

[craig0303]

Do you understand friendly or basic fractions? like how 3/6=4/8=1/2 ?

Basically once you get an answer to a fraction problem you want to put in lowest terms.
Say you have 8/12. If we express that in terms of primes it is 2*2*2/2/2/3. Two of the twos that are multiplied cancel with 2 that are divided and you end up with 2/3.

Alternately if you have something 15/3=5*3/3. The 3's cancel and you are left with 5.

but sometimes when you find a common denominator for all terms of both sides of the equal sign then the denominators dissapear

very frustrating having holes in my understanding in some basic math, but it happens, knowledge fades sometimes

 

[yenchin]

but sometimes when you find a common denominator for all terms of both sides of the equal sign then the denominators dissapear

very frustrating having holes in my understanding in some basic math, but it happens, knowledge fades sometimes

You said this many times. Could you please give us *explicit examples* of what you mean?

 

[craig0303]

i don't have one atm unfortunately, ahwell thanks anyways

 

[yenchin]

i don't have one atm unfortunately, ahwell thanks anyways

Well I am sure you are doing some study and exercises, go try them. Once you find such examples that confuse you, please come back and we will be here to help :-)

 

[arildno]

Well, perhaps you mean something like this:

$$\frac{1}{2}+ \frac{1}{4}+ \frac{2}{8}= \frac{4}{8}+ \frac{2}{8}+ \frac{2}{8}= \frac{8}{8}=1$$
Is that what you meant?

 

[craig0303]

Well, perhaps you mean something like this:

$$\frac{1}{2}+ \frac{1}{4}+ \frac{2}{8}= \frac{4}{8}+ \frac{2}{8}+ \frac{2}{8}= \frac{8}{8}=1$$
Is that what you meant?

nope :( wish i had a sample, just something that came to mind that i don't elaborately understand but am able to do, very frustrating

thanks all for helping, ill let you know if i run across a good sample of a time when you don't need the denominator once all terms have been given the same denominator and equivalent effect is taken upon the numerator on both sides of the equal sign which negates the need for the denominator

suppose i should have posted my question a little more well prepared lol

 

[DivisionByZro]

nope :( wish i had a sample, just something that came to mind that i don't elaborately understand but am able to do, very frustrating

thanks all for helping, ill let you know if i run across a good sample of a time when you don't need the denominator once all terms have been given the same denominator and equivalent effect is taken upon the numerator on both sides of the equal sign which negates the need for the denominator

suppose i should have posted my question a little more well prepared lol

This is by far the most confusing post I've read in a while.

 

[arildno]

Might it be the following valid manipulation of an equation:

Line 1:
$$\frac{x-2}{6}+\frac{1}{6}=\frac{5+2x}{6}$$
Line 2:
$$x-2+1=5+2x$$

 

[craig0303]

Might it be the following valid manipulation of an equation:

Line 1:
$$\frac{x-2}{6}+\frac{1}{6}=\frac{5+2x}{6}$$
Line 2:
$$x-2+1=5+2x$$

ya, that sort of thing, why is it the denominator can be disregarded but on other equations it has to stay even when the whole equation is on same terms, just would like some explanation of the technicality of it, math theory of it i suppose

 

[micromass]

They just multiplied the entire equation by 6. For example

$$\frac{x}{6}=\frac{1}{6}$$

implies

$$6\frac{x}{6}=6\frac{1}{6}$$

hence

$$x=1$$

 

[yenchin]

ya, that sort of thing, why is it the denominator can be disregarded but on other equations it has to stay even when the whole equation is on same terms, just would like some explanation of the technicality of it, math theory of it i suppose

You might be confused about "equal" as opposed to "imply". We have 2/4 + 3/4 = 5/4 *implies* 2 +3 = 5 but certainly 2/4 + 3/4 is *not equal* to 2 + 3.

 

[arildno]

ya, that sort of thing, why is it the denominator can be disregarded but on other equations it has to stay even when the whole equation is on same terms, just would like some explanation of the technicality of it, math theory of it i suppose

OK, I think I understand what bother you.

In the follwing case:
$$\frac{2}{5}+\frac{1}{5}=\frac{3}{5}$$
you wonder why you can't "remove" the denominator, and be "left with" just 3.
Right?

The reason is quite simple; you CAN remove 5, but if you should do that, you must do it on BOTH sides of the equality, namely be left with the equality:
2+1=3 (which is, of course, correct)

Both in this case, and in the case of the equation I gave you, what you are doing is to change the quantity each side represents equally much (by multiplying each side with the same number).

And that is valid.

However, if you just want to SIMPLIFY an expression, but not change its value, then you should NOT multiply what you get with anything.

Furthermore:
Read closely what yenchin says about the crucial distinction between "imply" and "equal"

 

[craig0303]

You might be confused about "equal" as opposed to "imply". We have 2/4 + 3/4 = 5/4 *implies* 2 +3 = 5 but certainly 2/4 + 3/4 is *not equal* to 2 + 3.

hmm, i guess that is where my confusion is, i assumed that the "implied" was a real disappearance, so other than formula manipulation and division of fractions (1/3 to .333 for example) you can't make a fraction denominator really disappear, so even when you make a common denominator it is still always there, its just implied to be gone

if i have this straight and my minor confusion is cured, then than you all for the help in this one, much appreciated

 

[yenchin]

so even when you make a common denominator it is still always there, its just implied to be gone

It all depends on *what you want to do* as arildno said. If you want to simplified an equation, you don't then remove the common denominator, so for example if you want to simplify

x/2 + (3x)/4

then you find that you get

(2x)/4 + (3x)/4 = (5x)/4. That's all. Keep it as such. You *don't* then go ahead and make the denominator disappear! So it is "not implied to be gone", if that is what you meant.

 

[craig0303]

It all depends on *what you want to do* as arildno said. If you want to simplified an equation, you don't then remove the common denominator, so for example if you want to simplify

x/2 + (3x)/4

then you find that you get

(2x)/4 + (3x)/4 = (5x)/4. That's all. Keep it as such. You *don't* then go ahead and make the denominator disappear! So it is "not implied to be gone", if that is what you meant.

nope that's a different situation, i mean when you get a same denominator for each side of the equation then solve

 

[uart]

This thread is crazy. I've still got no idea of exactly what the OP is asking about.

 

[uart]

After reading this whole thread one more time I think that the OP can only be asking one of two possible things.

1. Why is it that the lowest denominator of a sum is sometimes smaller than that of the terms being added.

Like

$$\frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$$

Or

2. Maybe he's asking about equating numerators in an equation with equal denominators.

Like

$$\frac{5x + 3}{3} = \frac{7x - 1}{4}$$

$$\frac{20x + 12}{12} = \frac{21x - 3}{12}$$

$$20x + 12 = 21x - 3$$

$$x = 15$$

Is it the third line above where the denominator gets dropped from both sides of the equation that's confusing you Craig?

 

[craig0303]

$$\frac{5x + 3}{3} = \frac{7x - 1}{4}$$

$$\frac{20x + 12}{12} = \frac{21x - 3}{12}$$

$$20x + 12 = 21x - 3$$

$$x = 15$$

ya, that sort of problem, just wondering why the denominator gets dropped and is no longer written as a fraction, i know how to do the math, just understanding technicality of it eludes me, just would mind knowing the logic behind it, perhaps i was tired the day i learned that math lol i don't know.

 

[HallsofIvy]

once you get
$$\frac{20x+ 12}{12}= \frac{21x- 3}{12}$$
You can multiply both sides of the equation by 12.

 

[hillzagold]

The denominator disappears when it's 1. If your fraction is 20/4, you would divide the numerator and denominator by 4, and get 5/1, at which point you would write it as 5.

This is assuming you're not using decimals. If you are using decimals, you can switch from fractions to decimals whenever you want.

 

[craig0303]

thank you all for the help, I am not majorly into mathematics, pre-calculus is all I am up to and then ill be learning calculus afterwards, just trying to get a more elaborative understanding of basic math before i build on it, thank you.

 

[yenchin]

ya, that sort of problem, just wondering why the denominator gets dropped and is no longer written as a fraction, i know how to do the math, just understanding technicality of it eludes me, just would mind knowing the logic behind it, perhaps i was tired the day i learned that math lol i don't know.

The point is you are trying to solve for x; you can do whatever you want to the equations as long as you do the same operation on both sides of the equation. If you want to keep the denominator you can do so, but it is not helpful for our purpose. The final answer is still x= something.