# Lowest common denominator [Solved!]

**John** 19 Dec 2015, 06:53

### My question

I'm having a very difficult time trying to figure out what my lowest comman denominator is when adding or subracting complex fractions. I site the example:

a/6y -2b/3y4

### Relevant page

7. Addition and Subtraction of Fractions

### What I've done so far

Many examples!

X

I'm having a very difficult time trying to figure out what my lowest comman denominator is when adding or subracting complex fractions. I site the example:
a/6y -2b/3y4

Relevant page
<a href="/factoring-fractions/7-addition-subtraction-fractions.php">7. Addition and Subtraction of Fractions</a>
What I've done so far
Many examples!

## Re: Lowest common denominator

**Newton** 19 Dec 2015, 16:08

Hello John

[You are encouraged to use the math entry system for entering math. It makes it a lot easier for others to read! You are also encouraged to show us what you have done so we can see where you are getting stuck.]

When we are adding ordinary fraction (numbers only) we can find the

lowest common denominator like this:

Say we are adding `\frac{7}{18}` and `\frac{3}{20}`

We look at the prime factors of 18:

`18 = 2 \times 3 \times 3`

Prime factors of 20:

`20 = 2 \times 2 \times 5`

We've got two 3s in 18 which we need to include.

We've got two 2s in 20 which we need to include (we don't include the

2 from 18 because the two 2s already cater for that single 2)

We've got one 5 in 20 which we need to include.

So the lowest common denominator will be `3 \times 3 \times 2 \times 2 \times 5 = 180`.

In your example, which I presume means `\frac{a}{6y} -\frac{2b}{3y^4}`,

For `6y`, you've got

`6y = 2 \times 3 \times y`

Can you keep going for `3y^4` and finish it?

X

Hello John
[You are encouraged to use the math entry system for entering math. It makes it a lot easier for others to read! You are also encouraged to show us what you have done so we can see where you are getting stuck.]
When we are adding ordinary fraction (numbers only) we can find the
lowest common denominator like this:
Say we are adding `\frac{7}{18}` and `\frac{3}{20}`
We look at the prime factors of 18:
`18 = 2 \times 3 \times 3`
Prime factors of 20:
`20 = 2 \times 2 \times 5`
We've got two 3s in 18 which we need to include.
We've got two 2s in 20 which we need to include (we don't include the
2 from 18 because the two 2s already cater for that single 2)
We've got one 5 in 20 which we need to include.
So the lowest common denominator will be `3 \times 3 \times 2 \times 2 \times 5 = 180`.
In your example, which I presume means `\frac{a}{6y} -\frac{2b}{3y^4}`,
For `6y`, you've got
`6y = 2 \times 3 \times y`
Can you keep going for `3y^4` and finish it?

## Re: Lowest common denominator

**John** 20 Dec 2015, 02:57

so `3y^4=3 xx y xx y xx y xx y`

So is it going to be `6y^5`?

X

so `3y^4=3 xx y xx y xx y xx y`
So is it going to be `6y^5`?

## Re: Lowest common denominator

**Murray** 20 Dec 2015, 11:38

Not quite - you've double-counted one (only) of the `y`'s.

Can you finish the whole problem now?

X

Not quite - you've double-counted one (only) of the `y`'s.
Can you finish the whole problem now?

## Re: Lowest common denominator

**John** 21 Dec 2015, 11:39

So it's `6y^4`.

For the first fraction, we need to multiply `6y` by `y^3` to get `6y^4`, so also need to do the top: `(ay^3)/(6y^4)`

For the second fraction, need to multiply top and bottom by `2`: `(2 xx 2b)/(2 xx 3y^4) = (4b)/(6y^4)`

Doing the sum: `(ay^3)/(6y^4) - (4b)/(6y^4) = (ay^3 - 4b)/(6y^4)`

Does that look about right?

X

So it's `6y^4`.
For the first fraction, we need to multiply `6y` by `y^3` to get `6y^4`, so also need to do the top: `(ay^3)/(6y^4)`
For the second fraction, need to multiply top and bottom by `2`: `(2 xx 2b)/(2 xx 3y^4) = (4b)/(6y^4)`
Doing the sum: `(ay^3)/(6y^4) - (4b)/(6y^4) = (ay^3 - 4b)/(6y^4)`
Does that look about right?

## Re: Lowest common denominator

**Murray** 22 Dec 2015, 13:04

Yes, that's fine.

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