Subtract these two fractions:
n/(5 - n) and 3/(n^2 - 25)
2007-12-06 10:30:12 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
n/(5 - n) and 3/(n^2 - 25)
= n/(n-5) - 3/(n-5)(n+5)
n(n+5)- 3
--------------------
(n-5)(n+5)
2007-12-06 10:37:53 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Subtract these two fractions:
n/(5 - n) and 3/(n^2 - 25)
n / ( 5 - n ) = n ( 5 + n ) / ( 5 - n ) ( 5 + n )
3 / ( n ^ - 25 ) = 3 / ( 5 - n ) ( 5 + n )
so
n / ( 5 - n ) - 3 / ( 5 - n ) ( 5 + n )
n ( 5 + n ) / ( 5 - n )^2 - 3 / ( 5 - n ) ( 5 + n )
( n ( 5 + n ) - 3 ) / ( 5 - n ) ( 5 + n )
( 5n + n^2 - 3 ) / ( 5 - n ) ( 5 + n )
( n^2 + 5n - 3 ) / ( 5 - n ) ( 5 + n )
Numerator doesn't factor evenly so I'll leave it at that.
2007-12-06 18:39:37 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋
n/(5 - n) -3/(n^2 - 25)
n/(5 - n) =3/(n^2 - 25)
n*(n^2 - 25)=3*(5 - n)
n^3-25=15-3n
n^3+3n=15+25
n^3+3n=40
2007-12-06 18:37:14 · answer #3 · answered by judithgreanwood 3 · 0⤊ 2⤋
[08]
n/(5-n)-3/(n^2-25)
=n/(5-n) - 3/(n+5)(n-5)
=-n/(n-5) -3/(N+5)(n-5)
={-n(n+5)-3}/(n+5)(n-5)
=(-n^2-5n-3)/(n+5)(n-5)
2007-12-06 18:43:52 · answer #4 · answered by alpha 7 · 0⤊ 0⤋