Question

Simplify and give your answer in the form f(x)/g(x):

[(6/(x-1))-(1/(x+1))]/[(x/(x-1))+(1/(x+1))]

Answer 1

How far have you gotten with this problem?

Answer 2

Not very.

Answer 3

multiply top and bottom (carefully) by
\[(x-1)(x+1)\] to clear the fraction. that is probably easiest

Answer 4

It's best to break this into pieces. Start with just the stuff in the numerator. Simplify that. Then do the same with the denom.

Answer 5

you will get
\[6(x+1)-(x-1)\] in the numerator and
\[x(x+1)+(x-1)\] in the denominator. then so some algebra to multiply out and combine like terms. should be good from there

Answer 6

careful with the distributive law and so one, but now you have
\[\frac{6(x+1)-(x-1)}{x(x+1)+x-1}\] so that is most of the work

Answer 7

why do you multiply x(x+1) in the denom?

Answer 8

ok you have the denominator of
\[\frac{x}{x-1}+\frac{1}{x+1}\] yes?

Answer 9

unless i read it incorrectly. i mean this is your entire denominator.

Answer 10

so now when you multiply by
\[(x+1)(x-1)\] you get
\[(\frac{x}{x-1}+\frac{1}{x+1})(x+1)(x-1)\]

Answer 11

in the first term the x - 1 cancel giving
\[x(x+1)\]
and in the second term the x +1 cancel giving
\[1(x-1)\] so the whole denominator becomes
\[x(x+1)+x-1\]

Answer 12

of course you should then write this as
\[x^2+2x-1\]

Answer 13

ok. I get it now. You've been a great help.