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How to solve Simultaneous equation in Mathematica:

x^2 - (a^2 - 1)y^2 = 1 and y^2 - pz^2 = 1 where 0< a < 16 and 0 < y < 100 and 0 < p < 50 and 0 < z < 100.

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closed as off-topic by Daniel Lichtblau, MarcoB, rhermans, Sektor, halirutan Jun 1 '18 at 22:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Daniel Lichtblau, MarcoB, Sektor, halirutan
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ What have you tried? $\endgroup$ – Daniel Lichtblau Jun 1 '18 at 15:46
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    $\begingroup$ It's very discouraging to see questions that show little effort. If you expect to inspire anybody to volunteer their time to look at your problem, then start by doing your part. Read the Mathematica documentation. Get the Informed badge in the site by taking and understanding the tour. Please write your equations in Wolfram Mathematica language (Also in $\LaTeX$ if that helps). To add new information edit your question, use the comments section just for comments. $\endgroup$ – rhermans Jun 1 '18 at 19:00
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(Referring to the answer by @chuy in "Make Reduce produce nicer output").

Regarding a and p as parameters and x, y, z as variables.

red = Reduce[{x^2 - (a^2 - 1) y^2 == 1 , y^2 - p z^2 == 1, 0 < a < 16,
 0 < y < 100 , 0 < p < 50 , 0 < z < 100}, {x, y, z}, Reals];

TraditionalForm[
   red //. Or -> 
   Composition[(Column[#, Right, Background -> {{White, LightGray}}, 
   Frame -> All] &), List]]

enter image description here

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Read the documentations for Solve and Assuming. And understand the difference between Set(=) and Equal(==).

Assuming[
 {0 < a < 16,
  0 < y < 100,
  0 < p < 50,
  0 < z < 100},
 Solve[{
   x^2 - (a^2 - 1) y^2 == 1,
     y^2 - p z^2 == 1
   }]
 ]

$\left\{\\ \left\{p\to \frac{y^2-1}{z^2},x\to -\sqrt{a^2 y^2-y^2+1}\right\},\\ \left\{p\to \frac{y^2-1}{z^2},x\to \sqrt{a^2 y^2-y^2+1}\right\},\\ \{x\to -a,y\to -1,z\to 0\},\\ \{x\to a,y\to -1,z\to 0\},\\ \{x\to -a,y\to 1,z\to 0\},\\ \{x\to a,y\to 1,z\to 0\}\\ \right\}$

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    $\begingroup$ To @rhermans. I think, since the OP gives conditons for p and z , the pz is simply a typo. You should regard it as p*z . $\endgroup$ – Akku14 Jun 1 '18 at 19:37
  • $\begingroup$ @Akku14 thanks, I think you are right. It's corrected now. $\endgroup$ – rhermans Jun 1 '18 at 19:42

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