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I am new to this and was trying to solve an equation. Yet, it is taking more than 6 hours, and I am unsure if I typed something wrong or if my laptop settings are incompatible with the software. Any help would be appreciated!

So, firstly, I found the value of b and set b as one of the two outputs enter image description here

Solve[0==((-1)/(1-b+z b))-(r c/((1-z) (1-b)+(1-c) (z+b-z b)))-(r (1-z) (1-b) (2b+z-2-2z b)/((1-b+z b)((1-b)(1-z)+(1-c) (z+b-z b))^2)),b]

Then, I tried to find the value of z in terms of c, r, and b enter image description here

Solve[0==((-c (1-b))/((1-b) (1-z)+(1-c) (z-z b+b)))+((r b)/(1-b+b z))-((r (1-b)^2 (1-z) (2b-2b z-1))/((1-b+b z)((1-b) (1-z)+(1-c)(z-z b+b))^2)),z]

But I have not gotten any answer somehow after 6 hours of waiting. I am unsure what went wrong because I have just been using this software. Many thanks!

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    $\begingroup$ Please post code as text rather than as images. Otherwise, you're asking folks to type in all of your code from scratch which results in potential errors and fewer folks willing to help. $\endgroup$
    – JimB
    Commented Jan 12 at 18:59
  • $\begingroup$ @JimB Right, thanks for reminding! Sorry, I am totally new here! $\endgroup$
    – C. K
    Commented Jan 12 at 19:18
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    $\begingroup$ You have ugly nested radicals involving z in that second one. Unraveling it will be difficult for Solve I think. $\endgroup$
    – Daniel Lichtblau
    Commented Jan 12 at 19:26
  • $\begingroup$ @azerbajdzan Thank you for your reply! Does it mean the value of z will change the value of b, which then affect z's? My goal is to find the value of z in terms of c, r, and b, and I just wanted to replace b with the answer I got from Solve1. How should I do then? Thanks again $\endgroup$
    – C. K
    Commented Jan 12 at 19:29
  • $\begingroup$ @C. K: My previous comment was not correct, I deleted it. I overlook something... but see my answer. $\endgroup$
    – azerbajdzan
    Commented Jan 12 at 19:31

1 Answer 1

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Put the two equations in one Solve as a system of equations and use third argument as a variable to eliminate, i.e. b.

Solve[{0 == ((-c (1 - 
          b))/((1 - b) (1 - z) + (1 - c) (z - z b + b))) + ((r b)/(1 -
         b + b z)) - ((r (1 - b)^2 (1 - z) (2 b - 2 b z - 1))/((1 - 
          b + b z) ((1 - b) (1 - z) + (1 - c) (z - z b + b))^2)), 
  0 == ((-1)/(1 - b + 
        z b)) - (r c/((1 - z) (1 - b) + (1 - c) (z + b - 
            z b))) - (r (1 - z) (1 - 
        b) (2 b + z - 2 - 
         2 z b)/((1 - b + 
           z b) ((1 - b) (1 - z) + (1 - c) (z + b - z b))^2))}, z, b]

Manipulate[Plot3D[z /. %[[1]], {c, 0, 1}, {b, 0, 1}], {r, 0.1, 1}]
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  • $\begingroup$ Thank you so much!! $\endgroup$
    – C. K
    Commented Jan 12 at 19:35
  • $\begingroup$ I tried and seems like I got the answers I wanted, thanks!! I just have a follow-up question if you do not mind. So, the output gave me a few answers and I tried to set b as them and visualize it using Manipulate and Plot3D one by one. However, when I input Manipulate[Plot3D[z,{c, 0, 1}, {b, 0, 1}], {r, 0, 1}], it always shows a empty plot. But when I copy the outputs manually and replace z with it, it will work. May I know what I got wrong? @azerbajdzan $\endgroup$
    – C. K
    Commented Jan 12 at 20:03
  • $\begingroup$ @C. K: Maybe you did not assign any expression to z? $\endgroup$
    – azerbajdzan
    Commented Jan 12 at 20:15
  • $\begingroup$ Oh, I said it wrong. I set the answers I got as z (instead of b) one by one. Yet, it does not show anything. But when I copy and paste the answer (and replace z in the Plot3D order), I got a correct plot. $\endgroup$
    – C. K
    Commented Jan 12 at 20:25
  • $\begingroup$ @C. K: I do not see your code so I can not tell where is your problem, but I edited my answer with a code that works and it plots the first z. If you want to plot second z then use %[[2]]... $\endgroup$
    – azerbajdzan
    Commented Jan 12 at 21:03

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