Newtons Method to solve equations

Question

The question is the following: Use Newton’s method to solve $$x^3_1+x_2=1,\\ x^3_2−x_1=−1 $$ Indicate your initial condition and how many steps it requires to reach the tolerate of error to be within 10−6 .

I can generate the code for just one equation. I don't know how to do it for both equations. This is my code for one of them

 % Change here for different functions 
 f=@(x) x^(3)+x-1
 %this is the derivative of the above function
 df=@(x) 3*x^(2)+1
 % Change lower limit 'a' and upper limit 'b'
 a=0; b=1;
 x=a;
 for i=1:1:100
     x1=x-(f(x)/df(x));
     x=x1;
 end
 sol=x;
 fprintf('Approximate Root is %.15f',sol)
 a=0;b=1;
 x=a;
 er(5)=0;
 for i=1:1:5
     x1=x-(f(x)/df(x));
     x=x1;
     er(i)=x1-sol;
 end 
 plot(er) 
 xlabel('Number of iterations')
 ylabel('Error') 
 title('Error Vs. Number of iterations')

Answer 1

Your function is

$$\mathbf{f}(\mathbf{x}) = \begin{pmatrix} x_1^3+x_2 \\ -x_1 + x_2^3 \end{pmatrix}.$$

Compute its Jacobian. Solve

$$\mathbf{x}_{n+1} = \mathbf{x}_n - \mathbf{J}^{-1}(\mathbf{x}_{n})\mathbf{x}_n.$$

You can write the function in MATLAB syntax as

f = @(x)[x(1)^3+x(2); -x(1)+x(2)^3];

The Jacobian I will leave to you.