+0  
 
0
566
1
avatar

solve((x/y)*x=z, (x/z)*x=y, x^2=y + z) doesnt work. i guess its a lot to handle. i was trying to experiment around the fact that (sqrt(10)/2)*sqrt(10) = 5, but if you switched 5 and 2 it still equated properly. any thoughts on the matter?

 Feb 17, 2015

Best Answer 

 #1
avatar+118724 
+5

solve((x/y)*x=z, (x/z)*x=y, x^2=y + z) doesnt work. i guess its a lot to handle. i was trying to experiment around the fact that (sqrt(10)/2)*sqrt(10) = 5, but if you switched 5 and 2 it still equated properly. any thoughts on the matter?

 

$$\\((x/y)*x=z, (x/z)*x=y, x^2=y + z)\\\\
x^2=yz\qquad x^2=y+z\\\\
y*z=y+z$$

 

there is an infinite number of solutions to this

The only integer solutions are y=z=0   which is not correct because you cannot divide by 0

and y=z=2      which means   x=4

 

Reference graph:

http://www.wolframalpha.com/input/?i=y*z%3Dy%2Bz

 Feb 17, 2015
 #1
avatar+118724 
+5
Best Answer

solve((x/y)*x=z, (x/z)*x=y, x^2=y + z) doesnt work. i guess its a lot to handle. i was trying to experiment around the fact that (sqrt(10)/2)*sqrt(10) = 5, but if you switched 5 and 2 it still equated properly. any thoughts on the matter?

 

$$\\((x/y)*x=z, (x/z)*x=y, x^2=y + z)\\\\
x^2=yz\qquad x^2=y+z\\\\
y*z=y+z$$

 

there is an infinite number of solutions to this

The only integer solutions are y=z=0   which is not correct because you cannot divide by 0

and y=z=2      which means   x=4

 

Reference graph:

http://www.wolframalpha.com/input/?i=y*z%3Dy%2Bz

Melody Feb 17, 2015

1 Online Users