# Expanding brackets with fractions

1. Feb 27, 2010

### Gringo123

I have no problem expanding brackets with fractions generally, unless the fraction contains an unknown variable, such as in the following expression:

m/4[6m - 8] + m/2[10m - 2]

I know that the answer is:
12/2m2 - 3m

..but I have no idea how to get to that. Can anyone help?

2. Feb 27, 2010

### tiny-tim

Hi Gringo123!

Don't leave out brackets!!

I think you mean:

[m/4][6m - 8] + [m/2][10m - 2]

= [13/2]m2 - 3m​

Does that help?

3. Feb 28, 2010

### Gringo123

Hello again Tim
Thanks again for helping out.
The fractions are definitiely not in brackets. However, I've probably confused everyone with a typo. The answer should be:

13/2 m2 - 3m

4. Feb 28, 2010

### Char. Limit

I guess you just multiply each term in the bracket by the term outside.

5. Feb 28, 2010

### tiny-tim

The fractions definitiely are in brackets.

If they're not, then the [6m - 8] and the [10m - 2] would be on the bottom, and you'd never get the answer given.

6. Feb 28, 2010

### Mentallic

I don't believe the brackets matter in this case. m/4(6m-8) is to be read as (m/4)(6m-8). If the second factor were to be in the denominator as you claim then it should be written as m/(4(6m-8)).
This is how it works in most older calculators that use parenthesis also.

Yes that's exactly how it's done!

$$\frac{m}{4}(6m-8)+\frac{m}{2}(10m-2)$$

$$=\frac{m}{4}(6m)-\frac{m}{4}(8)+\frac{m}{2}(10m)-\frac{m}{2}(2)$$

7. Feb 28, 2010

### tiny-tim

No it isn't!!

See eg http://en.wikipedia.org/wiki/BODMAS#The_standard_order_of_operations"
The vagaries of some computer programs don't affect the standard BODMAS rules for human notation!
But I'm not an older calculator! …

… I'm still fresh!

Last edited by a moderator: Apr 24, 2017
8. Feb 28, 2010

### Mentallic

Omg you're right!
I guess I've been misinterpreting 1/2x as (1/2)x this whole time! I always took that if you aren't to use parenthesis, then only the very next symbol is in the denominator, in this case, just the 2.

and what if we have 1/2xy. Do I interpret this as $$\frac{y}{2x}$$ or $$\frac{1}{2xy}$$? I'm guessing the latter, because of the new rules I have just learnt!

9. Feb 28, 2010

### tiny-tim

Yup! 1/2xy = 1/(2xy)