14.7: Solving Systems of Equations

Solving a Linear System

Consider the following 3 by 3 system:

\(4*x1 - 2*x2 + x3 = 7\)

\(x1 + x2 + 5*x3 = 10\)

\(-2*x1 + 3*x2 - x3 = 2\)

We can solve this system symbolically using solve.

syms x1 x2 x3
expr1 = 4*x1 - 2*x2 + x3 == 7;
expr2 = x1 + x2 + 5*x3 == 10;
expr3 = -2*x1 + 3*x2 - x3 == 2;

solution = solve(expr1, expr2, expr3)

x1_solution = solution.x1
x2_solution = solution.x2
x3_solution = solution.x3

numeric_solution = double([solution.x1 solution.x2 solution.x3])

The result is returned as a structure. Each field of the structure stores the solution for one unknown variable. We use dot notation, such as solution.x1, to access each result:

solution =

struct with fields:

    x1: 124/41
    x2: 121/41
    x3: 33/41


x1_solution =

    124/41


x2_solution =

    121/41


x3_solution =

    33/41


numeric_solution =

    3.0244 2.9512 0.8049

Solving a Nonlinear System

Symbolic solve is not limited to linear equations. It can also solve many nonlinear systems.

For example, consider the following system:

\(x^2 + y^2 = 25\)

\(y = x + 1\)

syms x y
eq1 = x^2 + y^2 == 25;
eq2 = y == x + 1;
sol = solve(eq1, eq2, x, y)

x= double(sol.x)
double(sol.y)

This example finds the intersection points of a circle and a line. Since the line may cross the circle at more than one point, the solution can contain more than one value for x and y.

sol =

struct with fields:

    x: [2×1 sym]
    y: [2×1 sym]


x =

    -4
    3


y =

    -3
    4