MATLAB: Solving a multivariate equation in terms of one variable

algebraequation

So I have this long equation I'd like to solve in terms of one of its variables, t. Here is what I put in:
syms a b c d e f g h I j k l m n o t
solve((((((((1 - t)^3)*a + 3*((1 - t)^2)*t*d + 3*(1 - t)*(t^2)*g +
(t^3)*j) - m)*(3*((1 - t)^2)*(d - a) + 6*(1 - t)*t*(g - d) +
3*(t^2)*(j - g)) + ((((1 - t)^3)*b + 3*((1 - t)^2)*t*e + 3*(1 -
t)*(t^2)*h + (t^3)*k) - n)*(3*((1 - t)^2)*(e - b) + 6*(1 - t)*t*(h -
e) + 3*(t^2)*(k - h)) + ((((1 - t)^3)*c + 3*((1 - t)^2)*t*f + 3*(1 -
t)*(t^2)*i + (t^3)*l) - o)*(3*((1 - t)^2)*(f - c) + 6*(1 - t)*t*(i -
f) + 3*(t^2)*(l - i))) / ((((m - (((1 - t)^3)*a + 3*((1 - t)^2)*t*d +
3*(1 - t)*(t^2)*g + (t^3)*j))^2) + (n - (((1 - t)^3)*b + 3*((1 -
t)^2)*t*e + 3*(1 - t)*(t^2)*h + (t^3)*k))^2 + (o - (((1 - t)^3)*c +
3*((1 - t)^2)*t*f + 3*(1 - t)*(t^2)*i + (t^3)*l))^2)^.5)) + 2t - 1 ==
0, t)
All it gives me back is "Error: Unexpected MATLAB expression." I double checked my syntax and tested the same sort of expression with smaller equations and it worked fine. Is there a maximum equation length? If not, what am I doing wrong?

Best Answer

In your last line at the end you have:
... + 2t - 1 == ...
should be:
... + 2*t - 1 == ...
Fix that, and it runs. (I didn’t wait for it to finish.)
The full code:
syms a b c d e f g h I j k l m n o t
solve(((((((1 - t)^3)*a + 3*((1 - t)^2)*t*d + 3*(1 - t)*(t^2)*g + ...
(t^3)*j) - m)*(3*((1 - t)^2)*(d - a) + 6*(1 - t)*t*(g - d) + ...
3*(t^2)*(j - g)) + ((((1 - t)^3)*b + 3*((1 - t)^2)*t*e + 3*(1 - ...
t)*(t^2)*h + (t^3)*k) - n)*(3*((1 - t)^2)*(e - b) + 6*(1 - t)*t*(h - ...
e) + 3*(t^2)*(k - h)) + ((((1 - t)^3)*c + 3*((1 - t)^2)*t*f + 3*(1 - ...
t)*(t^2)*i + (t^3)*l) - o)*(3*((1 - t)^2)*(f - c) + 6*(1 - t)*t*(i - ...
f) + 3*(t^2)*(l - i))) / ((((m - (((1 - t)^3)*a + 3*((1 - t)^2)*t*d + ...
3*(1 - t)*(t^2)*g + (t^3)*j))^2) + (n - (((1 - t)^3)*b + 3*((1 - ...
t)^2)*t*e + 3*(1 - t)*(t^2)*h + (t^3)*k))^2 + (o - (((1 - t)^3)*c + ...
3*((1 - t)^2)*t*f + 3*(1 - t)*(t^2)*i + (t^3)*l))^2)^.5)) + 2*t - 1 == ...
0, t)
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