Because your system contains two redundant unknowns is easy to check that it has infinitely many solutions in four-dimensional space. But you do not look for solutions in the whole space, and on a discrete finite set  a=0.01 ... 7.11 by 0.01  and so on. I think that there does not exist exact solutions on this set.

 Natural formulation of the problem: on a given set of numbers to find a quartet  [a, b, c, d] , for which the left-hand sides of the system have the least differencies from the right-hand sides. As a measure of the difference consider  (7.11-(a+b+c+d))^2+(7.11-a*b*c*d)^2 . The procedure consists of two stages. To speed up work on the first stage we consider the wider step=0.1, and step=0.01 on the second stage near the optimum point of the  first stage.

restart:

R:=[infinity, []]:  # First stage

for a from 0.1 by 0.1 to 7.11 do

for b from a by 0.1 to 7.11 do

for c from b by 0.1 to 7.11 do

for d from c by 0.1 to 7.11 do

delta:=(7.11-(a+b+c+d))^2+(7.11-a*b*c*d)^2;

if delta<0.01 and delta<R[1] then R:=[delta, [a, b, c, d]]: fi:

od: od: od: od:

R;

a0:=R[2,1]: b0:=R[2,2]: c0:=R[2,3]: d0:=R[2,4]:  # Second stage

for a from a0-0.09 by 0.01 to a0+0.09 do

for b from b0-0.09 by 0.01 to b0+0.09 do

for c from c0-0.09 by 0.01 to c0+0.09 do

for d from d0-0.09 by 0.01 to d0+0.09 do

delta:=(7.11-(a+b+c+d))^2+(7.11-a*b*c*d)^2;

if delta<R[1] then R:=[delta, [a, b, c, d]]: fi:

od: od: od: od:

R;

         [.11296e-3, [.8, 1.8, 1.9, 2.6]]  # Result of the first stage

  [.266342400e-7, [.80, 1.76, 1.92, 2.63]]  # Result of the second stage

Verification:

7.11-convert([.80, 1.76, 1.92, 2.63],`+`);

7.11-convert([.80, 1.76, 1.92, 2.63],`*`);

                                 0.

                         0.00016320

fa:=proc(x)

local L, i, j;

L:=[];

for i while x[i]=0 do

od;

L:=[op(L), i];

for j from -1 by -1 while x[j]=0 do

od;

[op(L), ArrayNumElems(x)+j+1];

end proc; 

Example:

fa(<1|0|0|1|0|0>);

          [1, 4]

It is easy to prove that in any interval  RealRange(n, n+1) , where  n is positive integer, function  x -> frac(x*floor(x)) - 1/2  has exactly n roots. Thereforу in  RealRange(1,100)  there are  add(n, n=1..99) = 4950  roots.  

Use geometry package. Specify the coordinates of all the objects. The solution is obvious:

assume(y1>0, x2>0, x3>0, x3<x2):

point(A,0,0), point(B,x1,y1), point(C,x2,0), point(R,x3,0):

line(BC, [B, C]), line(AB, [A, B]), line(AC, [A, C]), ParallelLine(PR, R, BC):

intersection(P, PR, AB):

ParallelLine(PQ, P, AC):

intersection(Q, PQ, BC):

line(AQ, [A, Q]), line(BR, [B, R]):

intersection(M, PR, AQ), intersection(N, PQ, BR):

triangle(APM, [A, P, M]), triangle(PBN, [P, B, N]), triangle(RQC, [R, Q, C]):

S1, S2, S3:=area(APM), area(PBN), area(RQC);

is(S1+S2=S3);

3^((x+3)/(5*x-2))-4 >= 5*3^((9*x-7)/(5*x-2)):  #   The original inequality

convert((x+3)/(5*x-2), parfrac);

convert((9*x-7)/(5*x-2), parfrac);

# Change in the original inequality y=3^(17/5/(5*x-2))

solve(5*(3^(9/5))/y<=(3^(1/5))*y-4, y);  # Solution for y

solve(3^(17/5/(5*x-2))>=3*3^(4/5));  # The final result

restart;

with(RealDomain):

sys := {x = expand(a*(3*cos(t) - cos(3*t))), y = simplify(expand(a*(3*sin(t) - sin(3*t))))};

subs(sin(t)=solve(sys[2], sin(t)), subs(cos(t)=sqrt(1-sin(t)^2), sys[1]));

eq:=subs(y=abs(y), simplify(lhs(%)^2=rhs(%)^2)) ;

a:=1:   plots[implicitplot](eq, x=-4..4, y=-4..4, thickness=2, numpoints=10000);

The next procedure solves the problem:

Divs:=proc(n)

local L, M, Div, k, i, K, j, T, P;

uses numtheory;

if n=1 then return [1,1] else

L[1]:=[[n]];

M:=[];

Div:=divisors(n) minus {1, n};

for k in Div while n/k>=k do

M:=[op(M), [n/k, k]];

od;

L[2]:=M;

for i from 3 to bigomega(n) do

K:=[];

for j from 1 to nops(L[i-1]) do

Div:=divisors(L[i-1][j,1]) minus {1, L[i-1][j,1]};

for k in Div while L[i-1][j,1]/k>=k and k>=L[i-1][j,2] do

K:=[op(K),[L[i-1][j, 1]/k, k, op(L[i-1][j][2..nops(L[i-1][j])])]];

od; od;

L[i]:=K;

od;

L:=[seq(L[i], i=1..bigomega(n))];

M:=[]; T:=[seq(ithprime(k), k=1..bigomega(n))];

for i from 1 to nops(L) do

P[i]:=[seq(T[k], k=1..i)];

for j from 1 to nops(L[i]) do

M:=[op(M), mul(P[i][t]^(L[i][j,t]-1), t=1..nops(P[i]))];

od: od:

[n, min(M)]; fi;

end proc:

Example:

[seq(Divs(n), n=1..100)];

alpha:=Pi/2:

plot3d([r*cos(t), r*sin(t)*cos(alpha), r*sin(t)*sin(alpha)], t=0..Pi, r=0..1, scaling=constrained, axes=normal, view=[-1..1, -1..1, -1..1]);

Easy to prove that the formula

a(n)=-1/3*n^5+65/12*n^4-193/6*n^3+1057/12*n^2-108*n+48

gives the integer values ​​for each n.

The first 10 terms of the sequence:

1, 3, 6, 12, 33, 51, -22, -384, -1383, -3557

It  was questioned and answered. See http://www.mapleprimes.com/questions/130373-Let-A-Sequence-An---Defined-By-An136123351and

In Maple a function can be specified as follows:

 f:=x->f(x)

For your example:

f:=x->x^2+2*x:

f(x+1);

(x+1)^2+2*(x+1)

The natural domain and the range of the function  x->sqrt(x+1)  of real variable x  can be found as follows:

Domain:=solve(x+1>=0);

r:=convert(domain, list);

Range=minimize(sqrt(x+1), x=r[1]..r[2])..maximize(sqrt(x+1), x=r[1]..r[2]);

                                Domain:=RealRange(0, infinity)

                                            r:=[-1, infinity]

                                          Range=0..infinity

Thanks Aser. Your code is perfect!

Adjustment of Brian Bovril's  code:

restart;

with(combinat);

DartSum := proc (Darts, Total)

local L, N, S, x, n;

 L := choose([seq(10, i = 1 .. Darts), seq(20, i = 1 .. Darts), seq(30, i = 1 .. Darts), seq(40, i = 1 .. Darts), seq(50, i = 1 .. Darts)], Darts):

N := nops(L); n:=0;

 for x to N do

 S := add(i, i = L[x]);

if S <> Total

then next

else print(L[x]); n:=n+1;

 end if

 end do;

if n=0 then print("does not exist"); fi; 

 end proc;

The routine:

Darts:=proc(n::integer, L::list, Total::integer)  # n - the number of shots,  L -  the list of the numbers on the target,  Total - total amount of points  

local l, M, S, T, U, c;

l:=nops(L);

M:=[seq(floor(Total/L[i]), i=1..l)];

S:=seq([seq([k,L[i]], k=0..M[i])], i=1..l);

T:=combinat[cartprod]([S]):

U:=[];

while not T[finished] do

c:=T[nextvalue]();

if add(c[i,1], i=1..l)=n and add(c[i,1]*c[i,2], i=1..l)=Total then U:=[op(U), c]; fi;  od;

U; # List of all variants

end proc;

Examples:

We will use "a prime" sign instead "a dot" sign. Your second-order equation with the initial conditions  x''+2 x'+3 x=4 t, x(0)=1, x'(0)=2,  can be written as a system of first-order equations  {x'=y, y'=-3 x-2 y+4 t, x(0)=1, y(0)=2} . Then solve this system by  Euler method just as one first-order equation.

1) Find the equation of the tangent line at any point (x0, y0) of the circle.

2) Write down the condition that the tangent line passes through the point (4,1).

3) Solving the resulting system of two equations with two unknowns x0 and y0, find the point of contact, and then the  tangent line. Get two solutions.