Question

solve the equation 10/n^2-1+2n-5/n-1=2n+5/n+1

Answer 1

I'll assume the question is

\[\frac{10}{n^2 -1} + \frac{2n - 5}{n -1} = \frac{2n + 5}{n + 1}\]

eliminate the denominator by multiplying every term by (n-1)(n+1)

\[10 + (2n -5)(n+1) = (2n +5)(n -1)\]

simplify the expression

\[10 + 2n^2 -3n - 5 = 2n^2 + 3n - 5\]

collect like terms gives

\[10 - 6n = 0\]

so 6n = 10

n = 5/3

Answer 2

can you show how you eliminated the denominators?

Answer 3

ok... N^2 - 1 = (n-1)(n+1)

1st term
\[\frac{10}{(n -1)(n+1)} \times (n -1)(n+1) = 10\]

2nd term \[\frac{(2n - 5)}{n -1} \times (n-1)(n+1) = (2n -5)(n +1)\]

3rd term

\[\frac{(2n +5}{(n +1)} \times (n -1)(n +1) = (2n + 5)(n -1)\]

Answer 4

oooh thank you very much it now makes more sense

Answer 5

okay can you help me with this one 1/b+2+1/b+2=3/b+1

Answer 6

is this the problem..?
\[\frac{1}{b +2}+ \frac{1}{b +2} = \frac{3}{b +1}\]

Answer 7

yes

Answer 8

ok the left hand side can be simplifed

\[\frac{2}{b +2} = \frac{3}{b +1}\]

does that make sense

Answer 9

yea i got that part, so do you eliminate the denominator by multiplying by b+2

Answer 10

just cross multiply to eliminate the denominators

\[2(b +1) = 3(b +2)\]

Answer 11

really you multiply by (b+2)(b+1) but its the same as cross multiplying

Answer 12

oh okay thanks again