Shortly:

restart;
simplify(f*sqrt(g/f))  assuming g>=0,f>0;
is(sqrt(g)*sqrt(f)=f*sqrt(g/f))  assuming g>=0,f>0;

                                  sqrt(g)*sqrt(f)
                                         true

Download Ch1_3_Functions_Exam8_new.mw

eq[1] := -3*c[0]+3*c[1] = c[0]*(1-259*d[0]*(1/192)+43*d[1]*(1/64)-11*d[2]*(1/64)+(1/64)*d[3])-23/6;
eq[2] := -3*d[0]+3*d[1] = d[0]*(-2+23*c[0]*(1/80)+139*c[1]*(1/320)-17*c[2]*(1/160)+3*c[3]*(1/320))-1;
eq[3] := -4*c[0]*(1/3)+c[2]+(1/3)*c[3] = (8*c[0]*(1/27)+4*c[1]*(1/9)+2*c[2]*(1/9)+(1/27)*c[3])*(1-2657*d[0]*(1/5184)-343*d[1]*(1/1728)-185*d[2]*(1/1728)-79*d[3]*(1/5184))-3;
eq[4] := -4*d[0]*(1/3)+d[2]+(1/3)*d[3] = (8*d[0]*(1/27)+4*d[1]*(1/9)+2*d[2]*(1/9)+(1/27)*d[3])*(-2+109*c[0]*(1/8640)+553*c[1]*(1/1440)+559*c[2]*(1/2880)+37*c[3]*(1/1080))-5/6;
eq[5] := -(1/3)*c[0]-c[1]+4*c[3]*(1/3) = ((1/27)*c[0]+2*c[1]*(1/9)+4*c[2]*(1/9)+8*c[3]*(1/27))*(1-673*d[0]*(1/5184)-455*d[1]*(1/1728)-505*d[2]*(1/1728)-767*d[3]*(1/5184))-49/27;
eq[6] := -(1/3)*d[0]-d[1]+4*d[3]*(1/3) = ((1/27)*d[0]+2*d[1]*(1/9)+4*d[2]*(1/9)+8*d[3]*(1/27))*(-2-173*c[0]*(1/4320)+241*c[1]*(1/2880)+59*c[2]*(1/180)+2191*c[3]*(1/8640))-11/36;
eq[7] := c[0] = 1;
eq[8] := d[0] = 0;

Sol:=[solve({seq(eq[i],i=1..8)})]:
identify(evalf(Sol))[];  # 2 solutions

{c[0] = 1, c[1] = 0, c[2] = -1, c[3] = -2, d[0] = 0, d[1] = -1/3, d[2] = -1/3, d[3] = 0}, {c[0] = 1, c[1] = -2.330639950, c[2] = -22.8998219, c[3] = 66.2531085, d[0] = 0, d[1] = -1/3, d[2] = 29.42142564, d[3] = -120.1805217}
 

system1.mw

solve(-x+y^2=1, y);
`&+-`(%[1]);

It is easy to write a procedure that will do this automatically for any pair of roots that differ only in sign, but I do not see much point in this.

Edit. Nevertheless, I wrote a procedure that does the job automatically. This procedure works for equations with one unknown. If there are roots of multiplicity> 1, then the procedure does not combine them with the roots of the opposite sign, so as not to interfere with the perception of multiple roots:

Download DoPlusMinus.mw

restart;
A:=u[1, 0](t)*v[1, 3](t)+u[1, 1](t)*v[1, 2](t)+u[1, 2](t)*v[1, 1](t)+u[1, 3](t)*v[1, 0](t);
B:=map(p->op(1,p)*subs(t=tau,op(2,p)), A);

Applying  LinearAlgebra:-LinearSolve  command to your system, we get a message from Maple, that the system is inconsistent:

restart;

omega := v/h:
t := sum(a[j]*x^j, j = 0 .. 6)+a[7]*cos(omega*x)+a[8]*sin(omega*x):
r1 := diff(t, x$2):
r2 := diff(t, x$4):
c1 := eval(t, x = q+2*h) = y[n+2]:
c2 := eval(r1, x = q) = f[n]:
c3 := eval(r1, x = q+h) = f[n+1]:
c4 := eval(r1, x = q+2*h) = f[n+2]:
c5 := eval(r1, x = q+3*h) = f[n+3]:
c6 := eval(r2, x = q) = g[n]:
c7 := eval(r2, x = q+h) = g[n+1]:
c8 := eval(r2, x = q+2*h) = g[n+2]:
c9 := eval(r2, x = q+3*h) = g[n+3]:
b1 := seq(a[i], i = 0 .. 8):

A,B:=LinearAlgebra:-GenerateMatrix([c1, c2, c3, c4, c5, c6, c7, c8, c9],[b1]):
LinearAlgebra:-LinearSolve(A,B);

          Error, (in LinearAlgebra:-LinearSolve) inconsistent system

The roots of your equation coincide with the roots of the numerator, which is a cubic polynomial (we denote it as  f(x) ). Therefore, there is at least one real root. Since the value  f(0)=-73  is negative so we take  a=0 . It remains to find  b>0  that  f(b)>0 . This value is easy to find in the for-loop:

Download find-interval.mw

RootOf  is a way of representing the roots of an equation when explicit formulas are too cumbersome or do not exist at all. The  index=...  indicates the ordinal number of the root if there are several of these roots. To get the roots explicitly, you can use  allvalues  command. But since your equation also contains the parameter  l , in order to get all the roots in explicit form, you need to specify the parameter value:

A:=RootOf(6*_Z^3+(27+3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z^2+(3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2-9*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+90*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-18*l^4+6*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2-81+45*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z-324-3*l^8+l^10*RootOf(_Z^2*l^2+3*_Z^4-3)^2+108*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)-3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^6+sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^8*RootOf(_Z^2*l^2+3*_Z^4-3)^2-63*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+30*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2+45*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2+351*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-108*l^4, index = 1):

allvalues(RootOf(_Z^2*l^2+3*_Z^4-3));

eval(A, l=1);
allvalues(%);

restart;
f[a,alpha,beta,k];
f[ax,k];
f[c,0,d];

The numerical solution is better and more accurate than graphical one.
The wage  y  is a function of the sales  x .

# Solution a
# Setting of this function
y:=x->2000+0.1*x;
y(1480);
                   Output:                   2148.0

# Solution b
solve(y(x)=2225, x);
                   Output:                    2250.

This integral  (=1/24)  is equal to the volume of the following body:

plot3d(x*y, x=0..1, y=0..1-x, style=surface, filled);

Should be  sin(2*x)  instead of  sin2x . This is just a symbol . The general rule is that the argument of any function must be surrounded by parentheses.

You can immediately get the result in vector form using  LinearAlgebra:-LinearSolve  command as in the following example:

Sys:={3*x+5*y=21, -2*x+y=-1}:
LinearAlgebra:-LinearSolve(<3,5; -2,1>, <21,-1>);

We get the symbolic result containing the imaginary unit  I , but in fact the result is a real number:

Digits:=20:
int(x^5/(exp(x)-1)/(1-exp(-x)), x=0..1);
evalf(%);
fnormal(%,19);
simplify(%, zero);

We get the symbolic result containing the imaginary unit  I , but in fact this is a real number:

Digits:=20:
int(x^5/(exp(x)-1)/(1-exp(-x)), x=0..1);
evalf(%);
fnormal(%,19);
simplify(%,zero);