Solving system of Non-linear Equations graphically

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Problem Statement: I want to solve a set of equations graphically by plotting their graphs and looking at the interesection. I know how to solve a linear system of equations in 2D, like this.
%Simulataneous Equations
%3x1+2x2 =10
% 2x1 - x2 =2
%Graphical Solution
x= 0:0.1:10;
x1 = (1/3).*(18-2.*x);
x2= -2 + 2.*x;
plot(x,x1,'r')
hold on
plot (x,x2,'b');
grid on
xlabel('x1')
ylabel('x2')
Now I want to do a similar thing but for a set of equations with two variables instead of one. The equations are
I want them to plot like I could for above example and see their intersection. I know second equation is just a constant and hence i expected a plane passing through the value 5.
My attempt:
% Making an vector for input variable
x1_1= linspace(0,20);
y1_1=ones(1,100)*5;
%making a grid
[x1_1,y1_1] = meshgrid(x1_1,y1_1);
%The third equation of the set
z1_1= (1 + x1_1.*y1_1)./(x1_1+x1_1^2);
%making a vector for other plot
y1_2=ones(1,100)*5;
z1_2 = linspace(0,20);
%making a grid
[y1_2,z1_2] = meshgrid(y1_2,z1_2);
x1_2= 1+y1_2-z1_2;
% Plotting for 3D graph.
surf(x1_1,y1_1,z1_1);
hold on
surf(x1_2,y1_2,z1_2);
I did not do for the third equation but all I get are two lines instead of two planes intersecting at some point. If any help or a small hint with the code can be provided, i would be thankful to you.
Thank you.

Accepted Answer

Ameer Hamza
Ameer Hamza on 23 Apr 2020
try this
figure;
ax = axes();
view(3);
hold(ax);
% First equation
x1 = linspace(-20,20);
y1 = linspace(-20,20);
%making a grid
[x1,y1] = meshgrid(x1,y1);
%The third equation of the set
z1= 1 + y1 - x1;
surf(ax, x1, y1, z1, 20*ones(size(x1)));
% Second equation
x1 = linspace(-20,20);
z1 = linspace(-20,20);
[x1,z1] = meshgrid(x1,z1);
y1 = 5*ones(size(x1));
surf(ax, x1, y1, z1, 10*ones(size(x1)));
% Third equation
x1 = linspace(-20,20);
y1 = linspace(-20,20);
%making a grid
[x1,y1] = meshgrid(x1,y1);
%The third equation of the set
z1= (1 + x1.*y1)./(x1+x1.^2);
surf(ax, x1, y1, z1, 1*ones(size(x1)));
zlim([-20 20])
legend({'Eq1', 'Eq2', 'Eq3'})
  6 Comments
Vira Roy
Vira Roy on 25 Apr 2020
I managed to get the correct graph from your help Ameer. Just a final remark. can I know the intersection point as well.
Thanks...

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