# What's the best way to check if a system of one equation and one inequality has any solution?

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Richárd Tóth on 9 Aug 2019
Commented: Torsten on 9 Aug 2019
Hello
I've tried the solve function, but it gave me empty result:
syms x y
>> cond1 = x^2/25 + y^2/25 == 1;
>> cond2 = y <= x-4;
>> conds = [cond1 cond2];
>> sol = solve(conds, [x y], 'ReturnConditions', true);
Tried it on wolframalpha which shows that there is solution : https://www.wolframalpha.com/input/?i=x%5E2%2F25+%2B+y%5E2%2F25+%3D+1+and+y+%3C%3D+x-4
So what do you recommend if I'm not interested in the solutions only want to know if there is any solution or not?

Torsten on 9 Aug 2019
Use "fmincon" with an arbitrary objective function (e.g. 0).

Richárd Tóth on 9 Aug 2019
nice, if exitflag=-2 that means that there is no solution, otherwise there is at least one solution, right?
Torsten on 9 Aug 2019
Usually yes.
To be sure, just evaluate "nonlcon" for the solution fmincon returns.
ceq should be 0 and c should be <=0.

Bruno Luong on 9 Aug 2019
Edited: Bruno Luong on 9 Aug 2019
In your case use first Euler-Lagrange condition gives one solution if it exists
meaning find lambda, x, y, st
[2*x/25, 2*y/25] = lambda * [-1,1]
x^2/25 + y^2/25 = 1
That gives 2 solutuions
(x,y) = sqrt(1/2)*5*[1,-1];
and
(x,y) = sqrt(1/2)*5*[-1,1];
Then check for the inequalities. The first solution meets the inequalities
y <= x-4
This solution does not require any toolbox or numerical calculation. Just pencil and paper. If you take some care with Kuhn Turker condition with the sign of lambda, you can even discard one of the two solutions before the last step.