how do you solve. these of fractions... 5/n-3 + n/n+3
Combine them and their denominator (5+n)/(n-3) Done lol
really lol
lol....no
It's not that simple.
yea i seen it was wrong lol i think i may have got it tho. just a lot of steps but after working about 5 i think i get it
He doesn't even explain what happened to the n + 3
yea i know
\[\frac{5}{n - 3} + \frac{3}{n + 3}\]
To add fractions like these, let's go back to actual numbers. Suppose you had to add \[\frac{2}{3} + \frac{4}{5}\] What would you do?
find the lcd
Oh sorry i didn't see it was n-3 and n+3 i thought they were both n-3 sorry
This chapter was annoying man are you on flvs?
You would realize that you need to multiply the first fraction by 5/5 and the second fraction by 3/3 \[\frac{2}{3} \times \frac{5}{5} + \frac{4}{5} \times \frac{3}{3}\] We do this because in order to add fractions we need both denominators to be the same.
\[[(n+3)/(n+3)]5/(n-3) + n/(n+3)[(n-3)/(n-3)]\]
Upon doing so we get \[\frac{10}{15} + \frac{12}{15}\] Which equals \[\frac{22}{15}\]
The same process can be applied to the fractions you have to add.
\[\frac{ 5n+15+n^2-3n }{ (n+3)(n-3) }\]
\[\frac{5}{n - 3} \times \frac{n + 3}{n + 3} + \frac{n}{n + 3} \times \frac{n - 3}{n - 3}\]
\[\frac{ n^2+2n+15 }{ (n+3)(n-3) }\]
\[\frac{ (n+5)(n-3) }{ (n+3)(n-3) }\]
cancel out the n-3's
\[\frac{ n+5 }{ n+3 }\]
that's the answer
Get ITTTT ?
Only thing is, it's still an improper fraction
I know but in the textbook, anywhere that will be the answer
#adding and subtracting rational expressions chapter 7 section 3 XDD
I was going to finish by saying that \[\frac{22}{15}\] is an improper fraction. And you express the improper fraction in simpliest from as a mixed fraction: \[\frac{22}{15} = \frac{15 + 7}{15} = \frac{15}{15} + \frac{7}{15} = 1 \frac{7}{15}\]
\[\frac{n + 5}{n + 3}\] is also an improper fraction. To express it as a mixed fraction, do the following: \[\frac{n + 5}{n + 3} = \frac{n + 3 + 2}{n + 3} = \frac{n + 3}{n + 3} + \frac{2}{n + 3} = 1 + \frac{2}{n + 3}\]
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