[quote author=Teaser4]when u wana request for bedspace u will login to the portal,with ur jamb registration no,and ur birth year as ur password...u gona change it later. First u will pay all the required fees in to oau bank account.after then u will come to skul(resumption) u can pay on campus if u wish.cos those banks may not confirm ur payments in time,if ur payments have not been confirmed u cant do anything on the portal,not even bedspace clearance...that's why I said it is slow nd stresssful,it depend

when r we to pay d fees n pls any info on payment of acceptance fee?

[quote author=Ghettoguru]

Question 1,

x can only be solved in terms of y and z, same way y can only be solved in terms of x and z; and z in terms of x and y.

1/x + 1/y + 1/z = 1

Let's solve for x in terms of y and z>

(yz + xz + xy)/xyz = 1
yz + xz + xy = xyz....equation (1)
xyz - xy - xz = yz
x(yz - y - z) = yz
x = yz/[yz - y - z]
x = yz/[y(z - 1) - z]

Now, we can solve for y in terms of x and z from equation (1)

yz + xz + xy = xyz
xyz - yz - xy = xz
y(yz - z - x) = xz
y = xz/(yz - x - z)
y = xz/

thanks i solved it that way b4 but the guy that gave me the question said it has real value solutions. Just wanted to b sure.

i dnt seem to get the second question the guy said the ans is a=256 which is true when i substituted it into the question.
can it b solved?
i got something like this
5(loga base 4)square + 48=a/8 loga base 4.

Solve
1. 1/X + 1/y + 1/z = 1 find x,y and z


2. 5Loga base 4 + 48log4 base a = a/8.

When r we resuming

nice.one:
wow! i really missed out on this maths frenzy maths is so fun that it became a past time back then in school. right now am just wallowing in nostalgia of the abstract/complex maths i did during my MSc in control systems engineering.i would love to be part of this but right now am way 2busy. i hail all the guru that is making things happen. maybe i will be recommending you guys for the Melina maths prize .
just hv these few contribution to make
for;
i) X^2=2^X
apart from numerical solution (graphical, etc) it can also be solved by differentiating twice and solving for X

ii) 3^X +2^X=5^x this can also be solved by playing around diff.

ans x=4 or x=2

2nioshine: @doubleDX once again gr8 wuk....any who can should try this...2^x=4x.... question2
x+y+z=2, x^2+y^2+z^2=26,
x^3+y^3+z^3=38....plz no graphical soln..ans=4....Q2ans=4,-3 &1

question1 has 2 solutions: 4 and 0.3099

ultranet: pls can u solve dis question: 3^x=9x...(interested in d working)

solution x=3 and x=0.128

ultranet: pls can u solve dis question: 3^x=9x...(interested in d working)

solution x=3 and x=0.128

Maths gurus pls help me solve the simultaneous equations

x + y =5
x^x + y^y=31