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Matt J
on 11 Mar 2014

Edited: Matt J
on 11 Mar 2014

Given a fixed value of x, you can find one corresponding y at which the expression is zero as follows

x=3;

f=@(y) x.^2+y.^3-10;

y_zero = fzero(f, y_guess);

Although, when f() is a polynomial, as in your example, it would be more efficient to use ROOTS instead of FZERO. Also, ROOTS will find you all solutions y for that x, instead of just one.

Similarly for given y, you can find one associated x as follows

y=1;

f=@(x) x.^2+y.^3-10;

x_zero = fzero(f, x_guess);

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