How to use fsolve with input from an array

Hi, I wrote down a simple code, I want to solve it by "fsolve" but with applying an array of values instead of one value for "t", I do not know how I can do it, please help. Thanks in advance.

function y=func(x)
y(1)=3*x(1)+x(2)-t;
y(2)=x(1)-2*x(2)-3;
end

t is an array of [4 5 6]

Actually I simplified my main problem. So, this is my main code:

function F = func(x)
tetha=0.2;
F=zeros(3,1);
F(1,1) = x(1)+(tetha/(1-tetha))*x(2)-(0.21/(1-tetha));
F(2,1) = x(3)-(x(2)*(1-x(1))/(x(1)*(1-x(2))));
F(3,1) = x(3)-(2.98*((x(1)*(x(3)-1)+1-0.1*x(3))/(x(1)*(x(3)-1)+1)-0.1));
end 

And here it is its solver:

clear
clc
fun = @func;
x0 = [0.5,0.5,1];
x(:,1)=fsolve(fun,x0);
disp(x)

It is ok with one value for tetha, but the problem is that tetha is an array of [0.2 0.4 0.6 0.8]. In fact, I need this system of equations to be solved 4 times (length of tetha) for x(1), x(2) , and x(3) and for each time output be displayed.

[KALYAN ACHARJYA]

function y=func(x)
y(1)=3*x(1)+x(2)-t;
y(2)=x(1)-2*x(2)-3;
end

t is not there in the input argument list??

y=func(x)

Or sorry I did not understand the question properly, can you restructure the question again, y(2) independent with t ??

[Amirreza Reyhanifar]

Actually I simplified my main problem. So, this is my main code:

function F = func(x)
tetha=0.2;
F=zeros(3,1);
F(1,1) = x(1)+(tetha/(1-tetha))*x(2)-(0.21/(1-tetha));
F(2,1) = x(3)-(x(2)*(1-x(1))/(x(1)*(1-x(2))));
F(3,1) = x(3)-(2.98*((x(1)*(x(3)-1)+1-0.1*x(3))/(x(1)*(x(3)-1)+1)-0.1));
end 

And here it is its solver:

clear
clc
fun = @func;
x0 = [0.5,0.5,1];
x(:,1)=fsolve(fun,x0);
disp(x)

It is ok with one value for tetha, but the problem is that tetha is an array of [0.2 0.4 0.6 0.8]. In fact, I need this system of equations to be solved 4 times (length of tetha) for x(1), x(2) , and x(3) and for each time output be displayed.

Answer 1

[Alan Stevens]

Presumably you want to find x(1) and x(2) given values for y(1) and y(2). If so, then you can simply solve them as follows:

t = [4 5 6];
M = [3 1; 1 -2];
y0 = [6; 7]; % replace with your own values

for i = 1:3
    
    y = y0 + [t(i); 3];
    x(:,i) = M\y;
    
end

giving results like

x =

4.2857 4.5714 4.8571

-2.8571 -2.7143 -2.5714

where each column gives x(1) and x(2) for the various values of t.

[Amirreza Reyhanifar]

Thanks for your answer, but I need it to be solved with fsolve