equation solver solve() returns empty result

I tried using symbolic equation solver to get the results

syms V a b K
eq1 = pi^2*(a+b).*(b-a).^2 == V
eq2 = V == 8000
eq3 = a == K*b
eq4 = K == 0.2:0.1:0.7
a = solve([eq1,eq2,eq3,eq4],a)

And I got

a =
 
Empty sym: 0-by-1

It should have solutions while it returned nothing. Did I have something wrong?

[madhan ravi]

why not solve the first three equations and substitute K values?

[kairui wang]

Honestly I'm try learning solve(). I want to figure out what's wrong with the codes.

[Walter Roberson]

eq4 = K == 0.2:0.1:0.7

defines a system of 6 equations,

[ K == 1/5, K == 3/10, K == 2/5, K == 1/2, K == 3/5, K == 7/10]

all of which have to be true simultaneously in order for solve() to find a solution.

You need to use symbolic K and substitute in the values later -- or you need to loop with specific numeric K values.

Also, you must solve() for as many variables as your provide equations. With four equations, you need to solve for four variables simultaneously. If you look at the solution I provided, I provide two equations and I solve for two variables.

Answer 1

[Walter Roberson]

V = 8000;
Kvals = 0.2:0.1:0.7;
syms a b K
eq1 = pi^2*(a+b).*(b-a).^2 == V
eq3 = a == K*b;
sol = solve([eq1, eq3], [a, b]);
a = subs(sol.a, K, Kvals)

You will find that the result is 3 (different solutions) per K value. Two of the solutions are complex valued and the third is real valued.

[kairui wang]

It seems there're complex solutions, is there a way to get sole real solutions?

[Walter Roberson]

At the end add

a(all(imag(a)~=0,2), :) = [];

This will delete all rows in which all of the values are complex-valued. If there should happen to be any row in which some of the values are only real and some are complex then this code will not delete such a row. The code could be changed to create a cell array in which each entry had 1 to 3 real-valued solutions (since the number of real-valued could potentially change according to k value).