@Muhammad Usman 

Expr:=(A^2/nu)^(1/3)*g*beta*q2[w]*sqrt(nu^2/A)*T(y, t)/((nu*A)^(1/3)*k)+(A^2/nu)^(1/3)*g*beta*T[infinity]/(nu*A)^(1/3)-(A^2/nu)^(1/3)*g*beta*V[infinity]/(nu*A)^(1/3)+(A^2/nu)^(1/3)*g*beta1*C(y, t)/(nu*A)^(1/3)-(A^2/nu)^(1/3)*g*beta1*C[infinity]/(nu*A)^(1/3);
simplify(Expr) assuming A>0, nu>0;
expand(%);

@torabi  I did not understand what this graph means.

If you want to plot  y(t) vs A  with a horizontal A-axis , then you need to go into 3-d space. You will have a surface such that its section with a plane  A=const  will be the corresponding plot of  y(t) :

restart;
P:=proc(A)
local omega, mu, B, eq1, eq2, eq3, dsys3;
omega := -2.667; mu := 10; B := 1;
eq1 := diff(x(t), t) = omega*x(t)-y(t)^2;
eq2 := diff(y(t), t) = mu*(z(t)-y(t));
eq3 := diff(z(t), t) = A*y(t)-B*z(t)+x(t)*y(t);
dsys3 := {eq1, eq2, eq3, x(0) = 10, y(0) = 10, z(0) = 10};
dsolve(dsys3, numeric);
end proc:

plot3d([(A,s)->A, (A,s)->s, (A,s)->eval(y(t),P(A)(s))], 1..10, 0..3, labels=[A,t,y(t)], labelfont=[times,16], numpoints=10000, axes=normal);

@torabi  In this case, what do you want to match A for each A?

@assma It is necessary to remove the square brackets and put the  multiplication sign where it was missed:
restart;
f:=(x,y)->exp(-0.96374054*x)*sin(2*Pi*y)*((-0.96374054)^2-4*Pi^2)/(2*Pi);
plots:-contourplot(f(x,y), x=-0.5..1.5, y=-0.5..1.5, numpoints=10000);

@Markiyan Hirnyk 

1. The same can be done simplier:

solve([x^3+x^2-y^2, -y+2*x+1], [x, y], explicit);

2.  I changed the head.

@Markiyan Hirnyk  @Carl Love  I do not understand why such intricate ways, when  RootFinding:-Isolate command  immediately gives the desired result:

restart;
with(RootFinding):
sol := Isolate([x^3+x^2-y^2, -y+2*x+1], [x, y]);    
Addition.  To solve this system, we can do without calling  RootFinding  package, applying  solve  command, but we will have to tinker to get rid of imaginary parts that are actually 0 :

restart;
sol := simplify(fnormal(evalf(solve([x^3+x^2-y^2, -y+2*x+1], [x, y], explicit))), zero);         

@mapleatha  In Maple 2015-2017 everything works properly. Execute  restart  command first:

restart;
subs([A=1/6,B=8/15,C=3/10], A/(s+1)+B/(s-2)+C/(s+3));

@max125 

1. Try yourself to find errors in your procedures.

2. Of course, this procedure can be written much shorter, for example:

Orthocenter:=proc(A::list, B::list, C::list)
local H;
if (B[1]-A[1])*(C[2]-A[2])=(B[2]-A[2])*(C[1]-A[1]) then error "The points A, B, C lie on one straight line" fi;
H:=proc(A, B, C)
local N;
N:=B-A;
N[1]*('x'-C[1])+N[2]*('y'-C[2])=0;
end proc;
solve({H(A,B,C), H(B,C,A)});
end proc:

Example of use:

Orthocenter([-3,3], [1,3], [-3,0]);

                                                  {x = -3, y = 3}

The problem is to symbolically prove that the determinant is 0. This is trivial:

LinearAlgebra:-Determinant(M):
is(%=0);
                                                              true

Another option:

simplify(LinearAlgebra:-Determinant(M));
                                                                0

@digerdiga  indexed  is the type of, for example,  a[n] :

whattype(a[n]);
                                  indexed

op(a[n]);
                                       n

The  evalindets  command  (see above)  applies  the procedure  t->op(t)*p+q+b[op(t)]  to any operand  t  of  eq1  of the  indexed  type. 

@Thomas Richard  Thank you very much! The third item on the list "Questions" helped.

I have the same problem in 32 bit Maple 2017.3 on Windows 10. The problem appeared about 2.5 months ago. Since then, I have reinstalled Maple and even been able to update it to Maple 2017.3, but the problem has remained (the standard GUI hangs on boot).

@pgl771  This is even simplier:

plot3d(y^2+x, x = 0 .. 2, y = x-2 .. x, orientation = [220, 75]);

@Les  For plotting in geom3d  package the  draw  command should be used:

draw(pp);

@Carl Love  The Iterator package was introduced in Maple 2016, therefore, unfortunately, OP will not be able to check the operation of your procedure.