Perform the indicated operation: 2/m-n^2 + 1/n^2-m
- Thread starter Helen
- Start date
Re: Perform the indicated operation.
Or better yet...
(2 / (m - n^2)) + (1 / (n^2 - m))
remember:
\(\displaystyle \L \frac{a}{b} + \frac{c}{d} = \frac{ad + cb}{bd}\)
So for your sum of two rationals:
\(\displaystyle \L \frac{2}{m - n^2} + \frac{1}{n^2 - m}\)
= \(\displaystyle \L \frac{2(n^2-m) + (m-n^2)}{(m-n^2)(n^2-m)}\)
= \(\displaystyle \L \frac{2n^2-2m + m-n^2}{(m-n^2)(n^2-m)}\)
= \(\displaystyle \L \frac{(n^2-m)}{(m-n^2)(n^2-m)}\)
...keep going. What can you now cancel? You will notice that your denominator will be (m-n^2)... you can switch that around so the largest term is first, and move that negative up to the numerator so it looks good.
Denis said:
Or better yet...
(2 / (m - n^2)) + (1 / (n^2 - m))
remember:
\(\displaystyle \L \frac{a}{b} + \frac{c}{d} = \frac{ad + cb}{bd}\)
So for your sum of two rationals:
\(\displaystyle \L \frac{2}{m - n^2} + \frac{1}{n^2 - m}\)
= \(\displaystyle \L \frac{2(n^2-m) + (m-n^2)}{(m-n^2)(n^2-m)}\)
= \(\displaystyle \L \frac{2n^2-2m + m-n^2}{(m-n^2)(n^2-m)}\)
= \(\displaystyle \L \frac{(n^2-m)}{(m-n^2)(n^2-m)}\)
...keep going. What can you now cancel? You will notice that your denominator will be (m-n^2)... you can switch that around so the largest term is first, and move that negative up to the numerator so it looks good.
D
Deleted member 4993
Guest
Helen said:
John is making the problem too complicated....
2/(m - n^2) + 1/(n^2 - m)
= 2/(m - n^2) - 1/(m - n^2)
= 1/(m - n^2)..............That's it
