Expanding brackets with fractions

[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?

 

[tiny-tim]

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?

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?

 

[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

 

[Char. Limit]

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

 

[tiny-tim]

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

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.

 

[Mentallic]

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?

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.

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

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)

 

[tiny-tim]

I don't believe the brackets matter in this case. m/4(6m-8) is to be read as (m/4)(6m-8).

No it isn't!

See eg http://en.wikipedia.org/wiki/BODMAS#The_standard_order_of_operations" …

Similarly, care must be exercised when using the slash ('/') symbol. The string of characters "1/2x" is interpreted by the above conventions as 1/(2x). If what is meant is (1/2) × x, then it should be written as (1/2)x. Again, the use of parentheses will clarify the meaning and should be used if there is any chance of misinterpretation.

The vagaries of some computer programs don't affect the standard BODMAS rules for human notation!

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.

But I'm not an older calculator! …

… I'm still fresh! ​

 

[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!

 

[tiny-tim]

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!

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