# Lowest common denominator [Solved!]

### 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

What I've done so far

Many examples!

Continues below

## Re: Lowest common denominator

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

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

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?

Continues below

## Re: Lowest common denominator

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)

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

Yes, that's fine.

X

Yes, that's fine.

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