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How to use Solve correctly?

Posted 10 years ago

Solve [ x1^2 + y1^2 == 6^2, x2^2 + y2^2 == 14^2, y2 == ((y1 - y2)/(x1 - x2))x2, 0 == ArcTan[((y2 - y1)/(x2 - x1) - y1/x1)/(1 + ((y2 - y1)/(x2 - x1))(y1/x1))], {x2, y2}]

Hi I'm a newbie to Solve...Can some one eyeball above an d let me know what I have configured incoreectly. 4 equations , 4 unknowns--should be able to solve yes ?

Regards,

John

POSTED BY: john massucci
2 Replies
Posted 10 years ago

Solve is often fine for polynomial problems and sometimes not aggressive enough for transcendental equations. And, if you look at the documentation closely, you might notice you were missing a set of { } around your list of equations. Reduce is sometimes better for transcendental problems, but gives solutions in a different form.

Simplify[ Reduce[{x1^2 + y1^2 == 6^2, x2^2 + y2^2 == 14^2, y2 == ((y1 - y2)/(x1 - x2))*x2, 
   0 == ArcTan[((y2-y1)/(x2 - x1) - y1/x1)/(1 + ((y2 - y1)/(x2 - x1))*(y1/x1))]}, {x2, y2}, Backsubstitution->True]]

yields, where && means And and || means Or

x1 != 0 && ((x2 == -((7 x1)/3) && ((Sqrt[36 - x1^2] + y1 == 0 && 7 Sqrt[36 - x1^2] == 3 y2) ||
 (Sqrt[36 - x1^2] == y1 && y2 == -(7/3) Sqrt[36 - x1^2]))) ||
 (7 x1 == 3 x2 && ((Sqrt[36 - x1^2] + y1 == 0 && y2 == -(7/3) Sqrt[36 - x1^2]) ||
 (Sqrt[36 - x1^2] == y1 && 7 Sqrt[36 - x1^2] == 3 y2))))

Removing that Backsubstitution->True will give you the results in a different form. Choose whichever better suits your needs.

POSTED BY: Bill Simpson

Your equations are not in a List.

In[6]:= Solve[{x1^2 + y1^2 == 6^2, x2^2 + y2^2 == 14^2, 
  y2 == ((y1 - y2)/(x1 - x2))*x2, 
  0 == ArcTan[((y2 - y1)/(x2 - x1) - 
       y1/x1)/(1 + ((y2 - y1)/(x2 - x1))*(y1/x1))]}, {x2, y2}]

Out[6]= {}

In[5]:= Reduce[{x1^2 + y1^2 == 6^2, x2^2 + y2^2 == 14^2, 
  y2 == ((y1 - y2)/(x1 - x2))*x2, 
  0 == ArcTan[((y2 - y1)/(x2 - x1) - 
       y1/x1)/(1 + ((y2 - y1)/(x2 - x1))*(y1/x1))]}, {x2, y2}]

Out[5]= x1 != 
  0 && (y1 == -Sqrt[36 - x1^2] || 
   y1 == Sqrt[36 - x1^2]) && (x2 == -((7 x1)/3) || x2 == (7 x1)/3) && 
 y2 == (x2 y1)/x1
POSTED BY: Frank Kampas
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