# S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
 It is currently Wed, 19 Jun 2013 03:33:56 UTC

 All times are UTC [ DST ]

 Page 1 of 1 [ 9 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: SystemPosted: Mon, 12 Mar 2012 20:28:30 UTC
 Senior Member

Joined: Sun, 25 Dec 2011 00:28:25 UTC
Posts: 95
Hello

I need help, solving the system.

Attempted:

I'm not really sure about what should I be doing with these kind of problems ,
I don't know if I started it right.. but I'd be delighted, if I could solve this one (with your help).. Thank you

Top

 Post subject: Re: SystemPosted: Mon, 12 Mar 2012 20:53:29 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12168
Location: Austin, TX
kreshnik wrote:
Hello

I need help, solving the system.

Attempted:

I'm not really sure about what should I be doing with these kind of problems ,
I don't know if I started it right.. but I'd be delighted, if I could solve this one (with your help).. Thank you

??

You have

and

If y=1 you get one thing, if not, cancel in the second given:

, and that's about as far as I can see it going. Your attempt somehow adds a square into the mix that isn't there in the statement of the problem. Either you made a typo in the problem statement and this is actually quite simple, or you did not and that's as far as you can go.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 06:16:37 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3724
Location: Ottawa Ontario
kreshnik wrote:
Attempted:

Good work; continue....since y^2-1 = (y-1)(y+1):

[(y+1)(y-1)]^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; since (y+1)^x = 100:

100(y-1)^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; cross multiply:

100(y+1)(y-1)^x = (y+1)(y-1)(y-1)^x ; do cancellations:

100 = y - 1
y = 101

(y+1)^x = 100
102^x = 100
x = LOG(100) / LOG(102) = .99571832568.....

_________________
Walked over to a beggar...he gave me a quarter...

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 06:22:59 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12168
Location: Austin, TX
Denis wrote:
kreshnik wrote:
Attempted:

Good work; continue....since y^2-1 = (y-1)(y+1):

[(y+1)(y-1)]^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; since (y+1)^x = 100:

100(y-1)^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; cross multiply:

100(y+1)(y-1)^x = (y+1)(y-1)(y-1)^x ; do cancellations:

100 = y - 1
y = 101

(y+1)^x = 100
102^x = 100
x = LOG(100) / LOG(102) = .99571832568.....

And what about the magical appearance of a square that was not there before?

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 06:35:26 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3724
Location: Ottawa Ontario
And what about the magical appearance of a square that was not there before?

Where's that? Can't see it! All I see is y^4 - 2y^2 + 1 = (y^2 - 1)^2, which is ok.
My solution:
y = 101
x = .99571832568.....
checks out if substituted in 2nd equation.

_________________
Walked over to a beggar...he gave me a quarter...

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 06:48:21 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12168
Location: Austin, TX
Denis wrote:
And what about the magical appearance of a square that was not there before?

Where's that? Can't see it! All I see is y^4 - 2y^2 + 1 = (y^2 - 1)^2, which is ok.
My solution:
y = 101
x = .99571832568.....
checks out if substituted in 2nd equation.

XD, yeah not quite sure why I thought it wasn't there.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 06:52:55 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3724
Location: Ottawa Ontario
T's'ok...1st mistake this year!

_________________
Walked over to a beggar...he gave me a quarter...

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 15:18:56 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3724
Location: Ottawa Ontario
kreshnik wrote:

Kreshnik, I find it often saves lots of time (and writing!) to go a bit like this:

Let a = y+1 and b = y-1 ; then a^x = 100 [1]

Since y^4 - 2y^2 + 1 = [(y+1)(y-1)]^2
then left side = (a^2 b^2)^(x-1) = [a^x b^x / (ab)]^2 [2]

right side = b^(2x) / a^2 = (b^x / a)^2 [3]

[2][3]: a^x b^x / (ab) = b^x / a
a^x b^x / b = b^x
[1] 100b^x / b = b^x
100b^x = (b)b^x
b = 100
y - 1 = 100
y = 101

Something for you to think about...

_________________
Walked over to a beggar...he gave me a quarter...

Top

 Post subject: Re: SystemPosted: Tue, 13 Mar 2012 18:47:55 UTC
 Senior Member

Joined: Sun, 25 Dec 2011 00:28:25 UTC
Posts: 95
Thank you everyone.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 9 posts ]

 All times are UTC [ DST ]

#### Who is online

Users browsing this forum: No registered users

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forum

Search for:
 Jump to:  Select a forum ------------------ High School and College Mathematics    Algebra    Geometry and Trigonometry    Calculus    Matrix Algebra    Differential Equations    Probability and Statistics    Proposed Problems Applications    Physics, Chemistry, Engineering, etc.    Computer Science    Math for Business and Economics Advanced Mathematics    Foundations    Algebra and Number Theory    Analysis and Topology    Applied Mathematics    Other Topics in Advanced Mathematics Other Topics    Administrator Announcements    Comments and Suggestions for S.O.S. Math    Posting Math Formulas with LaTeX    Miscellaneous