You have seen that fzero numerically finds where a function is zero in MATLAB. “Maple” uses the solve facility which can solve n simultaneous algebraic or transcendental equations for n unknowns. It firstly attempts to find an exact analytic solution. If this is not possible, it then attempts to find a numeric solution in variable precision format.

The command is of the form [a1,a2,...,an]=solve(f1,f2,...,fn,v1,v2,...,vn).

Here, f1,f2,...,fn are symbolic expressions that are to be made zero, v1,v2,...,vn are the variables in alphabetical order to be solved for, and a1,a2,...,an are the corresponding answers.

Solving a single equation

To solve a single equation f(x) = 0, this reduces to a=solve(f,x).

Of course there might be more than one solution for a. What happens if you do not specify the variable for which the equation is to be solved? If there is only one symbolic variable in the expression, it will solve for it by default. Otherwise x is the default variable. If there are two or more symbolic variables, none of which is x, and you forget to specify the one for which the solution is required, it appears to choose the last alphabetical one.

Example 1: Solve the equation

10/x² + 1 = 4− x. This is equivalent to solving 10/x² + 1 − 4 + x = 0. There are various ways:

clear all

syms x

f=10/(x^2+1)-4+x; % f is now a symbolic expression

soln=solve(f,x) % note the three solutions

or, knowing that x is the default variable,

clear all

syms x

soln=solve(10/(x^2+1)-4+x) % no need to use f or specify x

In earlier versions of MATLAB, you could use

clear all

soln=solve(’10/(x^2+1)=4-x’)

but this is being phased out.

 Exercise: Change to 10/(x^2+1)+2+x=0 and solve again.

Example 2: Solve the general quadratic equation ax² + bx + c = 0 for x.

clear all

syms a b c x

soln=solve(a*x^2+b*x+c)

giving the output soln=[1/2/a*(-b+(b^2-4*a*c)^(1/2))]

[1/2/a*(-b-(b^2-4*a*c)^(1/2))]

However, if for some reason you wanted to solve the same equation for b, you would need

clear all

syms a b c x

soln=solve(a*x^2+b*x+c,b)

giving the ouput soln=-(a*x^2+c)/x.

Example 3: We know sin θ = 0.5 has an infinite number of solutions. Enter

clear all

syms theta

angle=solve(sin(theta)-0.5)

giving the output angle=1/6*pi.

Example 4: Unlike Example 3, the equation sin θ = 0.5 − cos θ, −π ≤ θ ≤ π, does not have obvious solutions. But “Maple” can find two solutions exactly. The answers are hardly in a form that you would use and so it is better to convert them into numeric values.

clear all

syms theta

angle=solve(sin(theta)-0.5+cos(theta))   % hardly usable

angle=double(angle)                     % looks better in MATLAB

giving the output angle=[atan((1/4-1/4*7^(1/2))/(1/4+1/4*7^(1/2)))]

[atan((1/4+1/4*7^(1/2))/(1/4-1/4*7^(1/2)))+pi]

angle=-0.4240

1.9948

Example 5: By means of a simple sketch you can see that e^x = 4− x² has two solutions.

However, there is no analytic way to find them. Enter

clear all

syms x

soln=solve(exp(x)-4+x^2)

giving the output soln=1.0580064010906363086213874461232. This indicates that a numerical solution process was needed in “Maple”, and the answer is in variable precision format. If you wish, you can follow with soln=double(soln). Note that only the solution nearest to zero was found.

Solving simultaneous equations

Find the points of intersection of the two circles 2x² − x + 2y² − 8y = 0 and x² + 2x + y² − 6y + 1 = 0.

clear all                 % file available in M-files folder as file7.m

syms x y

eqn1=2*x^2-x+2*y^2-8*y;

eqn2=x^2+2*x+y^2-6*y+1;

[X,Y]=solve(eqn1,eqn2)

giving the output X = [ 2]                                   Y = [ 3]

                                  [ -14/41]                                  [ 3/41]

which implies the two points (x, y) = (2, 3) and (x, y) = (−14/41, 3/41).