DivideByZero Posted December 26, 2007 Share Posted December 26, 2007 A quick problem. Requires a little high school math reasoning. (x^2)(y) = x (z)(x) = y x, y, and z are not zero, +/-infinity, or +/- 1. Solve for x, y, and z. (there is a solution) Link to comment Share on other sites More sharing options...
Country Boy Posted December 26, 2007 Share Posted December 26, 2007 A quick problem. Requires a little high school math reasoning. (x^2)(y) = x (z)(x) = y x, y, and z are not zero, +/-infinity, or +/- 1. Solve for x, y, and z. (there is a solution) Since x is not 0, from the first equation y= 1/x. Then the second equation becomes zx= 1/x so z= 1/x^2. I'm not sure why you wrote "there is a solution". Generally, if you have 2 equations with 3 unknown numbers, there are an infinite number of solutions: choose x to be anything other than 0 or +/- 1 and solve for y and z. For example, x= 2, y= 1/2, z= 1/4 so x^2y= (4)(1/2)= 2= x, zx= (1/4)(2)= 1/2= y. Another, x=3, y= 1/3, z= 1/9 so that x^2y= 9(1/3)= 3= x, zx=(3)(1/9)= 1/3= y. Etc. Link to comment Share on other sites More sharing options...
Mr Skeptic Posted December 26, 2007 Share Posted December 26, 2007 A quick problem. Requires a little high school math reasoning. (x^2)(y) = x (z)(x) = y x, y, and z are not zero, +/-infinity, or +/- 1. Solve for x, y, and z. (there is a solution) Solution: x = 1/y y = y z = y^2 y may be anything but 0, +/- infinity, +/- 1. There are then an infinite number of solutions, as would be expected from 2 equations and 3 unknowns. Perhaps there is something missing here? Link to comment Share on other sites More sharing options...
Country Boy Posted December 27, 2007 Share Posted December 27, 2007 "DivideByZero" seems to think it amusing to give a question so vaguely phrased that he can then declare that no one has answered it correctly. Link to comment Share on other sites More sharing options...
DivideByZero Posted December 27, 2007 Author Share Posted December 27, 2007 My solution was actually [math]x=i^3[/MATH], [MATH]y=i[/MATH], [MATH]z=i^2[/MATH]. I was hoping this problem would only have one solution (this) but apparently I rushed myself to post this... Link to comment Share on other sites More sharing options...
thedarkshade Posted December 27, 2007 Share Posted December 27, 2007 "DivideByZero" seems to think it amusing to give a question so vaguely phrased that he can then declare that no one has answered it correctly. Yeah, you're right! And DivideByZero, this is not the place to ask question that you already know the answer so you can test others and make yourself look smart! This is a discussion forum! Link to comment Share on other sites More sharing options...
iNow Posted December 27, 2007 Share Posted December 27, 2007 Yeah, you're right! And DivideByZero, this is not the place to ask question that you already know the answer so you can test others and make yourself look smart! This is a discussion forum! darkshade, riddles are fine. Many members here enjoy a good riddle. It's just important that it make sense and is internally consistent. dividedbyzero, feel free to post another. Just make sure that it makes sense prior to doing so. Cheers. Link to comment Share on other sites More sharing options...
DivideByZero Posted December 27, 2007 Author Share Posted December 27, 2007 sorry thedarkshade, thanks iNow. Link to comment Share on other sites More sharing options...
thedarkshade Posted December 28, 2007 Share Posted December 28, 2007 darkshade, riddles are fine. Many members here enjoy a good riddle. It's just important that it make sense and is internally consistent. I think there is another place to post puzzles. It's called "Brain Teasers and Puzzles":mad: ! Link to comment Share on other sites More sharing options...
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