lol
Seriously though, why is OP ignoring everyone who offers to help? XD

kiddo your maths where off, start counting

because no one succeeded with the correct answers

you never even responded to me...what..lol

you didn't post anything here, so i call you

/troll

been a math tutor for over 4 years in pre-cal (professionally) just friekin post the questions and i can explain what to this of and how to solve each

really need these explained by tonite.... final exam on monday

finally posted,rofl.

@1: x^2+y=3 and x+y=1

from the second equation x+y=1 follows y=1-x,
pluck this is the first equation and you get

X^2+(1-x)=3 -> x^2-x-2=0 -> (x+1)(x-2)=0, so there are two solutions:

x=-1, y=2 and x=2, y=-1

@2: x+2y+1=0 and xy-y=0

here use the first equation and make it x=-2y-1
pluck into the other equation and you get

(-2y-1)y-y=0 -> -2y^2-y-y=0 -> -2y^2-2y=0 -> (divide -2) -> y^2+y=0 -> (y+1)y=0

so the two solutions are

y=-1, x=1 and y=0, x=-1

@5: (n+1)!/(2n) for n=1,2,3,4

(1+1)!/(2.1)=2!/2=2/2=1
(2+1)!/(2.2)=3!/4=6/4=3/2
(3+1)!/(2.3)=4!/6=24/6=4
(4+1)!/(2.5)=5!/8=120/8=15

@10: true, it's just 'renaming' n into k

@11: since all the variables are difference and there are (two different)(six different)(four different) there will be

2.6.4 different term, ie 48 terms

and since those terms come from three different sets each, each term will have 3 factors

This post was edited by brmv on Dec 28 2013 08:05pm